Practice · Product and sum signs fix how many are negative · Sensim Math

Variant 1 easy

Integers $a$ , $b$ , $c$ satisfy all three conditions below. Find the values of $a$ , $b$ and $c$ .

(a) $|a|<|b|<|c|$
(b) $a+b+c=7$
(c) $a\times b\times c=-12$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three integers with strictly increasing absolute values add to $7$ and multiply to $-12$ . Find them.

Givens

$|a| < |b| < |c|$
$a + b + c = 7$
$a \times b \times c = -12$

Unknowns

The values of $a$ , $b$ , $c$ .

Constraints

None of them is $0$ , since the product is $-12$ .
The three absolute values are different.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List#13 Convert to Algebra

Separate the signs from the sizes. The signs of the product and the sum decide how many of the numbers are negative; the absolute values must be three different whole numbers multiplying to $12$ , of which there are only two choices. Then the sum picks the one that works.

3 · Execute4 carry out the plan

1Count the negatives

#7 Identify Subproblems 7.NS.A.2

A negative product needs an odd number of negative factors:

1

or

3

. If all three were negative, the sum would be negative, not

7

. So exactly one is negative and two are positive.

abc = -12 < 0,\quad a + b + c = 7 > 0

Each negative factor flips the sign of the product.

2List the possible sizes

#2 Make a Systematic List 4.OA.B.4

|a|

,

|b|

,

|c|

are three different positive integers in increasing order with product

12

.

|a| \times |b| \times |c| = 12 = 1 \times 2 \times 6 = 1 \times 3 \times 4

Only

\{1, 2, 6\}

and

\{1, 3, 4\}

work;

1 \times 1 \times 12

and

2 \times 2 \times 3

repeat a value.

3Write the sum in terms of the negative one

#13 Convert to Algebra 7.NS.A.1

Let

q

be the absolute value of the negative number and

S

the total of the three absolute values. The other two are positive, so the sum is

(S - q) - q

.

(S - q) - q = S - 2q = 7

Subtracting the negative size from the two positive ones gives the sum.

4Test both size sets

#2 Make a Systematic List 7.NS.A.1

For

\{1, 2, 6\}

,

S = 9

and

q = 1

. For

\{1, 3, 4\}

,

S = 8

would need

q = \frac{1}{2}

, which is not an integer.

S = 9:\ 9 - 2q = 7,\ q = 1;\qquad S = 8:\ 8 - 2q = 7,\ q = \frac{1}{2}

So the negative number has size

1

and the positives have sizes

2

and

6

:

a = -1

,

b = 2

,

c = 6

.

Answer:

a=-1

,

b=2

,

c=6

4 · Reviewdoes it hold up?

$|-1| < |2| < |6|$ , $-1 + 2 + 6 = 7$ , and $-1 \times 2 \times 6 = -12$ . All three conditions hold.

Another way: Try every sign pattern directly: with sizes $1, 2, 6$ the sums are $-1+2+6 = 7$ , $1-2+6 = 5$ , $1+2-6 = -3$ ; with sizes $1, 3, 4$ they are $6$ , $2$ , $0$ . Only $-1+2+6$ reaches $7$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for products to count negative factors.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed integers to test which choice gives $7$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing factor triples of $12$.

💡Takeaway. Settle the signs first, then the sizes: two short lists beat one long search.

Variant 2 easy

Integers $a$ , $b$ , $c$ satisfy all three conditions below. Find the values of $a$ , $b$ and $c$ .

(a) $|a|<|b|<|c|$
(b) $a+b+c=-5$
(c) $a\times b\times c=12$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three integers with strictly increasing absolute values add to $-5$ and multiply to $12$ . Find them.

Givens

$|a| < |b| < |c|$
$a + b + c = -5$
$a \times b \times c = 12$

Unknowns

The values of $a$ , $b$ , $c$ .

Constraints

None of them is $0$ , since the product is $12$ .
The three absolute values are different.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List#13 Convert to Algebra

Separate the signs from the sizes. The signs of the product and the sum decide how many of the numbers are negative; the absolute values must be three different whole numbers multiplying to $12$ , of which there are only two choices. Then the sum picks the one that works.

3 · Execute4 carry out the plan

1Count the negatives

#7 Identify Subproblems 7.NS.A.2

A positive product needs an even number of negative factors:

0

or

2

. If none were negative, the sum would be positive, not

-5

. So exactly two are negative and one is positive.

abc = 12 > 0,\quad a + b + c = -5 < 0

Each negative factor flips the sign of the product.

2List the possible sizes

#2 Make a Systematic List 4.OA.B.4

|a|

,

|b|

,

|c|

are three different positive integers in increasing order with product

12

.

|a| \times |b| \times |c| = 12 = 1 \times 2 \times 6 = 1 \times 3 \times 4

Only

\{1, 2, 6\}

and

\{1, 3, 4\}

work;

1 \times 1 \times 12

and

2 \times 2 \times 3

repeat a value.

3Write the sum in terms of the positive one

#13 Convert to Algebra 7.NS.A.1

Let

p

be the absolute value of the positive number and

S

the total of the three absolute values. The other two are negative, so the sum is

p - (S - p)

.

p - (S - p) = 2p - S = -5

Subtracting the two negative sizes from the positive one gives the sum.

4Test both size sets

#2 Make a Systematic List 7.NS.A.1

For

\{1, 2, 6\}

,

S = 9

and

p = 2

. For

\{1, 3, 4\}

,

S = 8

would need

p = 1\frac{1}{2}

, which is not an integer.

S = 9:\ 2p - 9 = -5,\ p = 2;\qquad S = 8:\ 2p - 8 = -5,\ p = 1\frac{1}{2}

So the positive number has size

2

and the negatives have sizes

1

and

6

:

a = -1

,

b = 2

,

c = -6

.

Answer:

a=-1

,

b=2

,

c=-6

4 · Reviewdoes it hold up?

$|-1| < |2| < |-6|$ , $-1 + 2 + (-6) = -5$ , and $-1 \times 2 \times (-6) = 12$ . All three conditions hold.

Another way: Try every sign pattern directly: with sizes $1, 2, 6$ the sums are $-1-2+6 = 3$ , $-1+2-6 = -5$ , $1-2-6 = -7$ ; with sizes $1, 3, 4$ they are $0$ , $-2$ , $-6$ . Only $-1+2-6$ reaches $-5$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for products to count negative factors.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed integers to test which choice gives $-5$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing factor triples of $12$.

💡Takeaway. Settle the signs first, then the sizes: two short lists beat one long search.

Variant 3 easy

Integers $a$ , $b$ , $c$ satisfy all three conditions below. Find the values of $a$ , $b$ and $c$ .

(a) $|a|<|b|<|c|$
(b) $a+b+c=-7$
(c) $a\times b\times c=12$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three integers with strictly increasing absolute values add to $-7$ and multiply to $12$ . Find them.

Givens

$|a| < |b| < |c|$
$a + b + c = -7$
$a \times b \times c = 12$

Unknowns

The values of $a$ , $b$ , $c$ .

Constraints

None of them is $0$ , since the product is $12$ .
The three absolute values are different.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List#13 Convert to Algebra

Separate the signs from the sizes. The signs of the product and the sum decide how many of the numbers are negative; the absolute values must be three different whole numbers multiplying to $12$ , of which there are only two choices. Then the sum picks the one that works.

3 · Execute4 carry out the plan

1Count the negatives

#7 Identify Subproblems 7.NS.A.2

A positive product needs an even number of negative factors:

0

or

2

. If none were negative, the sum would be positive, not

-7

. So exactly two are negative and one is positive.

abc = 12 > 0,\quad a + b + c = -7 < 0

Each negative factor flips the sign of the product.

2List the possible sizes

#2 Make a Systematic List 4.OA.B.4

|a|

,

|b|

,

|c|

are three different positive integers in increasing order with product

12

.

|a| \times |b| \times |c| = 12 = 1 \times 2 \times 6 = 1 \times 3 \times 4

Only

\{1, 2, 6\}

and

\{1, 3, 4\}

work;

1 \times 1 \times 12

and

2 \times 2 \times 3

repeat a value.

3Write the sum in terms of the positive one

#13 Convert to Algebra 7.NS.A.1

Let

p

be the absolute value of the positive number and

S

the total of the three absolute values. The other two are negative, so the sum is

p - (S - p)

.

p - (S - p) = 2p - S = -7

Subtracting the two negative sizes from the positive one gives the sum.

4Test both size sets

#2 Make a Systematic List 7.NS.A.1

For

\{1, 2, 6\}

,

S = 9

and

p = 1

. For

\{1, 3, 4\}

,

S = 8

would need

p = \frac{1}{2}

, which is not an integer.

S = 9:\ 2p - 9 = -7,\ p = 1;\qquad S = 8:\ 2p - 8 = -7,\ p = \frac{1}{2}

So the positive number has size

1

and the negatives have sizes

2

and

6

:

a = 1

,

b = -2

,

c = -6

.

Answer:

a=1

,

b=-2

,

c=-6

4 · Reviewdoes it hold up?

$|1| < |-2| < |-6|$ , $1 + (-2) + (-6) = -7$ , and $1 \times (-2) \times (-6) = 12$ . All three conditions hold.

Another way: Try every sign pattern directly: with sizes $1, 2, 6$ the sums are $-1-2+6 = 3$ , $-1+2-6 = -5$ , $1-2-6 = -7$ ; with sizes $1, 3, 4$ they are $0$ , $-2$ , $-6$ . Only $1-2-6$ reaches $-7$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for products to count negative factors.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed integers to test which choice gives $-7$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing factor triples of $12$.

💡Takeaway. Settle the signs first, then the sizes: two short lists beat one long search.

Variant 4 medium

Integers $a$ , $b$ , $c$ satisfy all three conditions below. Find the values of $a$ , $b$ and $c$ .

(a) $|a|<|b|<|c|$
(b) $a+b+c=4$
(c) $a\times b\times c=-18$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three integers with strictly increasing absolute values add to $4$ and multiply to $-18$ . Find them.

Givens

$|a| < |b| < |c|$
$a + b + c = 4$
$a \times b \times c = -18$

Unknowns

The values of $a$ , $b$ , $c$ .

Constraints

None of them is $0$ , since the product is $-18$ .
The three absolute values are different.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List#13 Convert to Algebra

Separate the signs from the sizes. The signs of the product and the sum decide how many of the numbers are negative; the absolute values must be three different whole numbers multiplying to $18$ , of which there are only two choices. Then the sum picks the one that works.

3 · Execute4 carry out the plan

1Count the negatives

#7 Identify Subproblems 7.NS.A.2

A negative product needs an odd number of negative factors:

1

or

3

. If all three were negative, the sum would be negative, not

4

. So exactly one is negative and two are positive.

abc = -18 < 0,\quad a + b + c = 4 > 0

Each negative factor flips the sign of the product.

2List the possible sizes

#2 Make a Systematic List 4.OA.B.4

|a|

,

|b|

,

|c|

are three different positive integers in increasing order with product

18

.

|a| \times |b| \times |c| = 18 = 1 \times 2 \times 9 = 1 \times 3 \times 6

Only

\{1, 2, 9\}

and

\{1, 3, 6\}

work;

1 \times 1 \times 18

and

2 \times 3 \times 3

repeat a value.

3Write the sum in terms of the negative one

#13 Convert to Algebra 7.NS.A.1

Let

q

be the absolute value of the negative number and

S

the total of the three absolute values. The other two are positive, so the sum is

(S - q) - q

.

(S - q) - q = S - 2q = 4

Subtracting the negative size from the two positive ones gives the sum.

4Test both size sets

#2 Make a Systematic List 7.NS.A.1

For

\{1, 2, 9\}

,

S = 12

would need

q = 4

, which is not one of its sizes. For

\{1, 3, 6\}

,

S = 10

and

q = 3

.

S = 12:\ 12 - 2q = 4,\ q = 4;\qquad S = 10:\ 10 - 2q = 4,\ q = 3

So the negative number has size

3

and the positives have sizes

1

and

6

:

a = 1

,

b = -3

,

c = 6

.

Answer:

a=1

,

b=-3

,

c=6

4 · Reviewdoes it hold up?

$|1| < |-3| < |6|$ , $1 + (-3) + 6 = 4$ , and $1 \times (-3) \times 6 = -18$ . All three conditions hold.

Another way: Try every sign pattern directly: with sizes $1, 2, 9$ the sums are $-1+2+9 = 10$ , $1-2+9 = 8$ , $1+2-9 = -6$ ; with sizes $1, 3, 6$ they are $8$ , $4$ , $-2$ . Only $1-3+6$ reaches $4$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for products to count negative factors.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed integers to test which choice gives $4$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing factor triples of $18$.

💡Takeaway. Settle the signs first, then the sizes: two short lists beat one long search.

Variant 5 easy

Integers $a$ , $b$ , $c$ satisfy all three conditions below. Find the values of $a$ , $b$ and $c$ .

(a) $|a|<|b|<|c|$
(b) $a+b+c=-10$
(c) $a\times b\times c=18$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three integers with strictly increasing absolute values add to $-10$ and multiply to $18$ . Find them.

Givens

$|a| < |b| < |c|$
$a + b + c = -10$
$a \times b \times c = 18$

Unknowns

The values of $a$ , $b$ , $c$ .

Constraints

None of them is $0$ , since the product is $18$ .
The three absolute values are different.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List#13 Convert to Algebra

Separate the signs from the sizes. The signs of the product and the sum decide how many of the numbers are negative; the absolute values must be three different whole numbers multiplying to $18$ , of which there are only two choices. Then the sum picks the one that works.

3 · Execute4 carry out the plan

1Count the negatives

#7 Identify Subproblems 7.NS.A.2

A positive product needs an even number of negative factors:

0

or

2

. If none were negative, the sum would be positive, not

-10

. So exactly two are negative and one is positive.

abc = 18 > 0,\quad a + b + c = -10 < 0

Each negative factor flips the sign of the product.

2List the possible sizes

#2 Make a Systematic List 4.OA.B.4

|a|

,

|b|

,

|c|

are three different positive integers in increasing order with product

18

.

|a| \times |b| \times |c| = 18 = 1 \times 2 \times 9 = 1 \times 3 \times 6

Only

\{1, 2, 9\}

and

\{1, 3, 6\}

work;

1 \times 1 \times 18

and

2 \times 3 \times 3

repeat a value.

3Write the sum in terms of the positive one

#13 Convert to Algebra 7.NS.A.1

Let

p

be the absolute value of the positive number and

S

the total of the three absolute values. The other two are negative, so the sum is

p - (S - p)

.

p - (S - p) = 2p - S = -10

Subtracting the two negative sizes from the positive one gives the sum.

4Test both size sets

#2 Make a Systematic List 7.NS.A.1

For

\{1, 2, 9\}

,

S = 12

and

p = 1

. For

\{1, 3, 6\}

,

S = 10

would need

p = 0

, which is not one of its sizes.

S = 12:\ 2p - 12 = -10,\ p = 1;\qquad S = 10:\ 2p - 10 = -10,\ p = 0

So the positive number has size

1

and the negatives have sizes

2

and

9

:

a = 1

,

b = -2

,

c = -9

.

Answer:

a=1

,

b=-2

,

c=-9

4 · Reviewdoes it hold up?

$|1| < |-2| < |-9|$ , $1 + (-2) + (-9) = -10$ , and $1 \times (-2) \times (-9) = 18$ . All three conditions hold.

Another way: Try every sign pattern directly: with sizes $1, 2, 9$ the sums are $-1-2+9 = 6$ , $-1+2-9 = -8$ , $1-2-9 = -10$ ; with sizes $1, 3, 6$ they are $2$ , $-4$ , $-8$ . Only $1-2-9$ reaches $-10$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for products to count negative factors.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed integers to test which choice gives $-10$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing factor triples of $18$.

💡Takeaway. Settle the signs first, then the sizes: two short lists beat one long search.

Variant 6 medium

Integers $a$ , $b$ , $c$ satisfy all three conditions below. Find the values of $a$ , $b$ and $c$ .

(a) $|a|<|b|<|c|$
(b) $a+b+c=11$
(c) $a\times b\times c=-20$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three integers with strictly increasing absolute values add to $11$ and multiply to $-20$ . Find them.

Givens

$|a| < |b| < |c|$
$a + b + c = 11$
$a \times b \times c = -20$

Unknowns

The values of $a$ , $b$ , $c$ .

Constraints

None of them is $0$ , since the product is $-20$ .
The three absolute values are different.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List#13 Convert to Algebra

Separate the signs from the sizes. The signs of the product and the sum decide how many of the numbers are negative; the absolute values must be three different whole numbers multiplying to $20$ , of which there are only two choices. Then the sum picks the one that works.

3 · Execute4 carry out the plan

1Count the negatives

#7 Identify Subproblems 7.NS.A.2

A negative product needs an odd number of negative factors:

1

or

3

. If all three were negative, the sum would be negative, not

11

. So exactly one is negative and two are positive.

abc = -20 < 0,\quad a + b + c = 11 > 0

Each negative factor flips the sign of the product.

2List the possible sizes

#2 Make a Systematic List 4.OA.B.4

|a|

,

|b|

,

|c|

are three different positive integers in increasing order with product

20

.

|a| \times |b| \times |c| = 20 = 1 \times 2 \times 10 = 1 \times 4 \times 5

Only

\{1, 2, 10\}

and

\{1, 4, 5\}

work;

1 \times 1 \times 20

and

2 \times 2 \times 5

repeat a value.

3Write the sum in terms of the negative one

#13 Convert to Algebra 7.NS.A.1

Let

q

be the absolute value of the negative number and

S

the total of the three absolute values. The other two are positive, so the sum is

(S - q) - q

.

(S - q) - q = S - 2q = 11

Subtracting the negative size from the two positive ones gives the sum.

4Test both size sets

#2 Make a Systematic List 7.NS.A.1

For

\{1, 2, 10\}

,

S = 13

and

q = 1

. For

\{1, 4, 5\}

,

S = 10

would need

q = -\frac{1}{2}

, which is not an integer.

S = 13:\ 13 - 2q = 11,\ q = 1;\qquad S = 10:\ 10 - 2q = 11,\ q = -\frac{1}{2}

So the negative number has size

1

and the positives have sizes

2

and

10

:

a = -1

,

b = 2

,

c = 10

.

Answer:

a=-1

,

b=2

,

c=10

4 · Reviewdoes it hold up?

$|-1| < |2| < |10|$ , $-1 + 2 + 10 = 11$ , and $-1 \times 2 \times 10 = -20$ . All three conditions hold.

Another way: Try every sign pattern directly: with sizes $1, 2, 10$ the sums are $-1+2+10 = 11$ , $1-2+10 = 9$ , $1+2-10 = -7$ ; with sizes $1, 4, 5$ they are $8$ , $2$ , $0$ . Only $-1+2+10$ reaches $11$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for products to count negative factors.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed integers to test which choice gives $11$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing factor triples of $20$.

💡Takeaway. Settle the signs first, then the sizes: two short lists beat one long search.

Variant 7 medium

Integers $a$ , $b$ , $c$ satisfy all three conditions below. Find the values of $a$ , $b$ and $c$ .

(a) $|a|<|b|<|c|$
(b) $a+b+c=-9$
(c) $a\times b\times c=20$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three integers with strictly increasing absolute values add to $-9$ and multiply to $20$ . Find them.

Givens

$|a| < |b| < |c|$
$a + b + c = -9$
$a \times b \times c = 20$

Unknowns

The values of $a$ , $b$ , $c$ .

Constraints

None of them is $0$ , since the product is $20$ .
The three absolute values are different.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List#13 Convert to Algebra

Separate the signs from the sizes. The signs of the product and the sum decide how many of the numbers are negative; the absolute values must be three different whole numbers multiplying to $20$ , of which there are only two choices. Then the sum picks the one that works.

3 · Execute4 carry out the plan

1Count the negatives

#7 Identify Subproblems 7.NS.A.2

A positive product needs an even number of negative factors:

0

or

2

. If none were negative, the sum would be positive, not

-9

. So exactly two are negative and one is positive.

abc = 20 > 0,\quad a + b + c = -9 < 0

Each negative factor flips the sign of the product.

2List the possible sizes

#2 Make a Systematic List 4.OA.B.4

|a|

,

|b|

,

|c|

are three different positive integers in increasing order with product

20

.

|a| \times |b| \times |c| = 20 = 1 \times 2 \times 10 = 1 \times 4 \times 5

Only

\{1, 2, 10\}

and

\{1, 4, 5\}

work;

1 \times 1 \times 20

and

2 \times 2 \times 5

repeat a value.

3Write the sum in terms of the positive one

#13 Convert to Algebra 7.NS.A.1

Let

p

be the absolute value of the positive number and

S

the total of the three absolute values. The other two are negative, so the sum is

p - (S - p)

.

p - (S - p) = 2p - S = -9

Subtracting the two negative sizes from the positive one gives the sum.

4Test both size sets

#2 Make a Systematic List 7.NS.A.1

For

\{1, 2, 10\}

,

S = 13

and

p = 2

. For

\{1, 4, 5\}

,

S = 10

would need

p = \frac{1}{2}

, which is not an integer.

S = 13:\ 2p - 13 = -9,\ p = 2;\qquad S = 10:\ 2p - 10 = -9,\ p = \frac{1}{2}

So the positive number has size

2

and the negatives have sizes

1

and

10

:

a = -1

,

b = 2

,

c = -10

.

Answer:

a=-1

,

b=2

,

c=-10

4 · Reviewdoes it hold up?

$|-1| < |2| < |-10|$ , $-1 + 2 + (-10) = -9$ , and $-1 \times 2 \times (-10) = 20$ . All three conditions hold.

Another way: Try every sign pattern directly: with sizes $1, 2, 10$ the sums are $-1-2+10 = 7$ , $-1+2-10 = -9$ , $1-2-10 = -11$ ; with sizes $1, 4, 5$ they are $0$ , $-2$ , $-8$ . Only $-1+2-10$ reaches $-9$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for products to count negative factors.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed integers to test which choice gives $-9$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing factor triples of $20$.

💡Takeaway. Settle the signs first, then the sizes: two short lists beat one long search.

Variant 8 hard

Integers $a$ , $b$ , $c$ satisfy all three conditions below. Find the values of $a$ , $b$ and $c$ .

(a) $|a|<|b|<|c|$
(b) $a+b+c=5$
(c) $a\times b\times c=-24$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three integers with strictly increasing absolute values add to $5$ and multiply to $-24$ . Find them.

Givens

$|a| < |b| < |c|$
$a + b + c = 5$
$a \times b \times c = -24$

Unknowns

The values of $a$ , $b$ , $c$ .

Constraints

None of them is $0$ , since the product is $-24$ .
The three absolute values are different.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List#13 Convert to Algebra

Separate the signs from the sizes. The signs of the product and the sum decide how many of the numbers are negative; the absolute values must be three different whole numbers multiplying to $24$ , of which there are only four choices. Then the sum picks the one that works.

3 · Execute4 carry out the plan

1Count the negatives

#7 Identify Subproblems 7.NS.A.2

A negative product needs an odd number of negative factors:

1

or

3

. If all three were negative, the sum would be negative, not

5

. So exactly one is negative and two are positive.

abc = -24 < 0,\quad a + b + c = 5 > 0

Each negative factor flips the sign of the product.

2List the possible sizes

#2 Make a Systematic List 4.OA.B.4

|a|

,

|b|

,

|c|

are three different positive integers in increasing order with product

24

.

|a| \times |b| \times |c| = 24 = 1 \times 2 \times 12 = 1 \times 3 \times 8 = 1 \times 4 \times 6 = 2 \times 3 \times 4

Only

\{1, 2, 12\}

,

\{1, 3, 8\}

,

\{1, 4, 6\}

and

\{2, 3, 4\}

work;

1 \times 1 \times 24

and

2 \times 2 \times 6

repeat a value.

3Write the sum in terms of the negative one

#13 Convert to Algebra 7.NS.A.1

Let

q

be the absolute value of the negative number and

S

the total of the three absolute values. The other two are positive, so the sum is

(S - q) - q

.

(S - q) - q = S - 2q = 5

Subtracting the negative size from the two positive ones gives the sum.

4Test the size sets

#2 Make a Systematic List 7.NS.A.1

For

\{1, 2, 12\}

,

S = 15

would need

q = 5

, which is not one of its sizes. For

\{1, 3, 8\}

,

S = 12

would need

q = 3\frac{1}{2}

, which is not an integer. For

\{1, 4, 6\}

,

S = 11

would need

q = 3

, which is not one of its sizes. For

\{2, 3, 4\}

,

S = 9

and

q = 2

.

S = 15:\ 15 - 2q = 5,\ q = 5;\qquad S = 12:\ 12 - 2q = 5,\ q = 3\frac{1}{2};\qquad S = 11:\ 11 - 2q = 5,\ q = 3;\qquad S = 9:\ 9 - 2q = 5,\ q = 2

So the negative number has size

2

and the positives have sizes

3

and

4

:

a = -2

,

b = 3

,

c = 4

.

Answer:

a=-2

,

b=3

,

c=4

4 · Reviewdoes it hold up?

$|-2| < |3| < |4|$ , $-2 + 3 + 4 = 5$ , and $-2 \times 3 \times 4 = -24$ . All three conditions hold.

Another way: Try every sign pattern directly: with sizes $1, 2, 12$ the sums are $-1+2+12 = 13$ , $1-2+12 = 11$ , $1+2-12 = -9$ ; with sizes $1, 3, 8$ they are $10$ , $6$ , $-4$ ; with sizes $1, 4, 6$ they are $9$ , $3$ , $-1$ ; with sizes $2, 3, 4$ they are $5$ , $3$ , $1$ . Only $-2+3+4$ reaches $5$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for products to count negative factors.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed integers to test which choice gives $5$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing factor triples of $24$.

💡Takeaway. Settle the signs first, then the sizes: two short lists beat one long search.

Variant 9 medium

Integers $a$ , $b$ , $c$ satisfy all three conditions below. Find the values of $a$ , $b$ and $c$ .

(a) $|a|<|b|<|c|$
(b) $a+b+c=-13$
(c) $a\times b\times c=24$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three integers with strictly increasing absolute values add to $-13$ and multiply to $24$ . Find them.

Givens

$|a| < |b| < |c|$
$a + b + c = -13$
$a \times b \times c = 24$

Unknowns

The values of $a$ , $b$ , $c$ .

Constraints

None of them is $0$ , since the product is $24$ .
The three absolute values are different.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List#13 Convert to Algebra

Separate the signs from the sizes. The signs of the product and the sum decide how many of the numbers are negative; the absolute values must be three different whole numbers multiplying to $24$ , of which there are only four choices. Then the sum picks the one that works.

3 · Execute4 carry out the plan

1Count the negatives

#7 Identify Subproblems 7.NS.A.2

A positive product needs an even number of negative factors:

0

or

2

. If none were negative, the sum would be positive, not

-13

. So exactly two are negative and one is positive.

abc = 24 > 0,\quad a + b + c = -13 < 0

Each negative factor flips the sign of the product.

2List the possible sizes

#2 Make a Systematic List 4.OA.B.4

|a|

,

|b|

,

|c|

are three different positive integers in increasing order with product

24

.

|a| \times |b| \times |c| = 24 = 1 \times 2 \times 12 = 1 \times 3 \times 8 = 1 \times 4 \times 6 = 2 \times 3 \times 4

Only

\{1, 2, 12\}

,

\{1, 3, 8\}

,

\{1, 4, 6\}

and

\{2, 3, 4\}

work;

1 \times 1 \times 24

and

2 \times 2 \times 6

repeat a value.

3Write the sum in terms of the positive one

#13 Convert to Algebra 7.NS.A.1

Let

p

be the absolute value of the positive number and

S

the total of the three absolute values. The other two are negative, so the sum is

p - (S - p)

.

p - (S - p) = 2p - S = -13

Subtracting the two negative sizes from the positive one gives the sum.

4Test the size sets

#2 Make a Systematic List 7.NS.A.1

For

\{1, 2, 12\}

,

S = 15

and

p = 1

. For

\{1, 3, 8\}

,

S = 12

would need

p = -\frac{1}{2}

, which is not an integer. For

\{1, 4, 6\}

,

S = 11

would need

p = -1

, which is not one of its sizes. For

\{2, 3, 4\}

,

S = 9

would need

p = -2

, which is not one of its sizes.

S = 15:\ 2p - 15 = -13,\ p = 1;\qquad S = 12:\ 2p - 12 = -13,\ p = -\frac{1}{2};\qquad S = 11:\ 2p - 11 = -13,\ p = -1;\qquad S = 9:\ 2p - 9 = -13,\ p = -2

So the positive number has size

1

and the negatives have sizes

2

and

12

:

a = 1

,

b = -2

,

c = -12

.

Answer:

a=1

,

b=-2

,

c=-12

4 · Reviewdoes it hold up?

$|1| < |-2| < |-12|$ , $1 + (-2) + (-12) = -13$ , and $1 \times (-2) \times (-12) = 24$ . All three conditions hold.

Another way: Try every sign pattern directly: with sizes $1, 2, 12$ the sums are $-1-2+12 = 9$ , $-1+2-12 = -11$ , $1-2-12 = -13$ ; with sizes $1, 3, 8$ they are $4$ , $-6$ , $-10$ ; with sizes $1, 4, 6$ they are $1$ , $-3$ , $-9$ ; with sizes $2, 3, 4$ they are $-1$ , $-3$ , $-5$ . Only $1-2-12$ reaches $-13$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for products to count negative factors.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed integers to test which choice gives $-13$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing factor triples of $24$.

💡Takeaway. Settle the signs first, then the sizes: two short lists beat one long search.

Variant 10 hard

Integers $a$ , $b$ , $c$ satisfy all three conditions below. Find the values of $a$ , $b$ and $c$ .

(a) $|a|<|b|<|c|$
(b) $a+b+c=12$
(c) $a\times b\times c=-30$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three integers with strictly increasing absolute values add to $12$ and multiply to $-30$ . Find them.

Givens

$|a| < |b| < |c|$
$a + b + c = 12$
$a \times b \times c = -30$

Unknowns

The values of $a$ , $b$ , $c$ .

Constraints

None of them is $0$ , since the product is $-30$ .
The three absolute values are different.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List#13 Convert to Algebra

Separate the signs from the sizes. The signs of the product and the sum decide how many of the numbers are negative; the absolute values must be three different whole numbers multiplying to $30$ , of which there are only four choices. Then the sum picks the one that works.

3 · Execute4 carry out the plan

1Count the negatives

#7 Identify Subproblems 7.NS.A.2

A negative product needs an odd number of negative factors:

1

or

3

. If all three were negative, the sum would be negative, not

12

. So exactly one is negative and two are positive.

abc = -30 < 0,\quad a + b + c = 12 > 0

Each negative factor flips the sign of the product.

2List the possible sizes

#2 Make a Systematic List 4.OA.B.4

|a|

,

|b|

,

|c|

are three different positive integers in increasing order with product

30

.

|a| \times |b| \times |c| = 30 = 1 \times 2 \times 15 = 1 \times 3 \times 10 = 1 \times 5 \times 6 = 2 \times 3 \times 5

Only

\{1, 2, 15\}

,

\{1, 3, 10\}

,

\{1, 5, 6\}

and

\{2, 3, 5\}

work;

1 \times 1 \times 30

repeats a value.

3Write the sum in terms of the negative one

#13 Convert to Algebra 7.NS.A.1

Let

q

be the absolute value of the negative number and

S

the total of the three absolute values. The other two are positive, so the sum is

(S - q) - q

.

(S - q) - q = S - 2q = 12

Subtracting the negative size from the two positive ones gives the sum.

4Test the size sets

#2 Make a Systematic List 7.NS.A.1

For

\{1, 2, 15\}

,

S = 18

would need

q = 3

, which is not one of its sizes. For

\{1, 3, 10\}

,

S = 14

and

q = 1

. For

\{1, 5, 6\}

,

S = 12

would need

q = 0

, which is not one of its sizes. For

\{2, 3, 5\}

,

S = 10

would need

q = -1

, which is not one of its sizes.

S = 18:\ 18 - 2q = 12,\ q = 3;\qquad S = 14:\ 14 - 2q = 12,\ q = 1;\qquad S = 12:\ 12 - 2q = 12,\ q = 0;\qquad S = 10:\ 10 - 2q = 12,\ q = -1

So the negative number has size

1

and the positives have sizes

3

and

10

:

a = -1

,

b = 3

,

c = 10

.

Answer:

a=-1

,

b=3

,

c=10

4 · Reviewdoes it hold up?

$|-1| < |3| < |10|$ , $-1 + 3 + 10 = 12$ , and $-1 \times 3 \times 10 = -30$ . All three conditions hold.

Another way: Try every sign pattern directly: with sizes $1, 2, 15$ the sums are $-1+2+15 = 16$ , $1-2+15 = 14$ , $1+2-15 = -12$ ; with sizes $1, 3, 10$ they are $12$ , $8$ , $-6$ ; with sizes $1, 5, 6$ they are $10$ , $2$ , $0$ ; with sizes $2, 3, 5$ they are $6$ , $4$ , $0$ . Only $-1+3+10$ reaches $12$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for products to count negative factors.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed integers to test which choice gives $12$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing factor triples of $30$.

💡Takeaway. Settle the signs first, then the sizes: two short lists beat one long search.

Variant 11 hard

Integers $a$ , $b$ , $c$ satisfy all three conditions below. Find the values of $a$ , $b$ and $c$ .

(a) $|a|<|b|<|c|$
(b) $a+b+c=-16$
(c) $a\times b\times c=30$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three integers with strictly increasing absolute values add to $-16$ and multiply to $30$ . Find them.

Givens

$|a| < |b| < |c|$
$a + b + c = -16$
$a \times b \times c = 30$

Unknowns

The values of $a$ , $b$ , $c$ .

Constraints

None of them is $0$ , since the product is $30$ .
The three absolute values are different.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List#13 Convert to Algebra

Separate the signs from the sizes. The signs of the product and the sum decide how many of the numbers are negative; the absolute values must be three different whole numbers multiplying to $30$ , of which there are only four choices. Then the sum picks the one that works.

3 · Execute4 carry out the plan

1Count the negatives

#7 Identify Subproblems 7.NS.A.2

A positive product needs an even number of negative factors:

0

or

2

. If none were negative, the sum would be positive, not

-16

. So exactly two are negative and one is positive.

abc = 30 > 0,\quad a + b + c = -16 < 0

Each negative factor flips the sign of the product.

2List the possible sizes

#2 Make a Systematic List 4.OA.B.4

|a|

,

|b|

,

|c|

are three different positive integers in increasing order with product

30

.

|a| \times |b| \times |c| = 30 = 1 \times 2 \times 15 = 1 \times 3 \times 10 = 1 \times 5 \times 6 = 2 \times 3 \times 5

Only

\{1, 2, 15\}

,

\{1, 3, 10\}

,

\{1, 5, 6\}

and

\{2, 3, 5\}

work;

1 \times 1 \times 30

repeats a value.

3Write the sum in terms of the positive one

#13 Convert to Algebra 7.NS.A.1

Let

p

be the absolute value of the positive number and

S

the total of the three absolute values. The other two are negative, so the sum is

p - (S - p)

.

p - (S - p) = 2p - S = -16

Subtracting the two negative sizes from the positive one gives the sum.

4Test the size sets

#2 Make a Systematic List 7.NS.A.1

For

\{1, 2, 15\}

,

S = 18

and

p = 1

. For

\{1, 3, 10\}

,

S = 14

would need

p = -1

, which is not one of its sizes. For

\{1, 5, 6\}

,

S = 12

would need

p = -2

, which is not one of its sizes. For

\{2, 3, 5\}

,

S = 10

would need

p = -3

, which is not one of its sizes.

S = 18:\ 2p - 18 = -16,\ p = 1;\qquad S = 14:\ 2p - 14 = -16,\ p = -1;\qquad S = 12:\ 2p - 12 = -16,\ p = -2;\qquad S = 10:\ 2p - 10 = -16,\ p = -3

So the positive number has size

1

and the negatives have sizes

2

and

15

:

a = 1

,

b = -2

,

c = -15

.

Answer:

a=1

,

b=-2

,

c=-15

4 · Reviewdoes it hold up?

$|1| < |-2| < |-15|$ , $1 + (-2) + (-15) = -16$ , and $1 \times (-2) \times (-15) = 30$ . All three conditions hold.

Another way: Try every sign pattern directly: with sizes $1, 2, 15$ the sums are $-1-2+15 = 12$ , $-1+2-15 = -14$ , $1-2-15 = -16$ ; with sizes $1, 3, 10$ they are $6$ , $-8$ , $-12$ ; with sizes $1, 5, 6$ they are $0$ , $-2$ , $-10$ ; with sizes $2, 3, 5$ they are $0$ , $-4$ , $-6$ . Only $1-2-15$ reaches $-16$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for products to count negative factors.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed integers to test which choice gives $-16$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing factor triples of $30$.

💡Takeaway. Settle the signs first, then the sizes: two short lists beat one long search.

Variant 12 hard

Integers $a$ , $b$ , $c$ satisfy all three conditions below. Find the values of $a$ , $b$ and $c$ .

(a) $|a|<|b|<|c|$
(b) $a+b+c=-11$
(c) $a\times b\times c=32$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three integers with strictly increasing absolute values add to $-11$ and multiply to $32$ . Find them.

Givens

$|a| < |b| < |c|$
$a + b + c = -11$
$a \times b \times c = 32$

Unknowns

The values of $a$ , $b$ , $c$ .

Constraints

None of them is $0$ , since the product is $32$ .
The three absolute values are different.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List#13 Convert to Algebra

Separate the signs from the sizes. The signs of the product and the sum decide how many of the numbers are negative; the absolute values must be three different whole numbers multiplying to $32$ , of which there are only two choices. Then the sum picks the one that works.

3 · Execute4 carry out the plan

1Count the negatives

#7 Identify Subproblems 7.NS.A.2

A positive product needs an even number of negative factors:

0

or

2

. If none were negative, the sum would be positive, not

-11

. So exactly two are negative and one is positive.

abc = 32 > 0,\quad a + b + c = -11 < 0

Each negative factor flips the sign of the product.

2List the possible sizes

#2 Make a Systematic List 4.OA.B.4

|a|

,

|b|

,

|c|

are three different positive integers in increasing order with product

32

.

|a| \times |b| \times |c| = 32 = 1 \times 2 \times 16 = 1 \times 4 \times 8

Only

\{1, 2, 16\}

and

\{1, 4, 8\}

work;

1 \times 1 \times 32

,

2 \times 2 \times 8

and

2 \times 4 \times 4

repeat a value.

3Write the sum in terms of the positive one

#13 Convert to Algebra 7.NS.A.1

Let

p

be the absolute value of the positive number and

S

the total of the three absolute values. The other two are negative, so the sum is

p - (S - p)

.

p - (S - p) = 2p - S = -11

Subtracting the two negative sizes from the positive one gives the sum.

4Test both size sets

#2 Make a Systematic List 7.NS.A.1

For

\{1, 2, 16\}

,

S = 19

would need

p = 4

, which is not one of its sizes. For

\{1, 4, 8\}

,

S = 13

and

p = 1

.

S = 19:\ 2p - 19 = -11,\ p = 4;\qquad S = 13:\ 2p - 13 = -11,\ p = 1

So the positive number has size

1

and the negatives have sizes

4

and

8

:

a = 1

,

b = -4

,

c = -8

.

Answer:

a=1

,

b=-4

,

c=-8

4 · Reviewdoes it hold up?

$|1| < |-4| < |-8|$ , $1 + (-4) + (-8) = -11$ , and $1 \times (-4) \times (-8) = 32$ . All three conditions hold.

Another way: Try every sign pattern directly: with sizes $1, 2, 16$ the sums are $-1-2+16 = 13$ , $-1+2-16 = -15$ , $1-2-16 = -17$ ; with sizes $1, 4, 8$ they are $3$ , $-5$ , $-11$ . Only $1-4-8$ reaches $-11$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for products to count negative factors.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed integers to test which choice gives $-11$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing factor triples of $32$.

💡Takeaway. Settle the signs first, then the sizes: two short lists beat one long search.

Product and sum signs fix how many are negative · 12 practice problems

Generated variants — 12

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Where next?