Practice · Split by common divisors, count |b| values · Sensim Math

Variant 1 easy

How many ordered pairs $(a, b)$ of integers satisfy both of the following conditions?

(a) $\frac{4}{a}$ and $\frac{6}{a}$ are both integers.
(b) $\frac{1}{3}\le\frac{|b|}{a}<2$

Your answer

Show solution

1 · Understandwhat's really being asked

Count integer pairs $(a, b)$ where $a$ divides both $4$ and $6$ , and $\frac{|b|}{a}$ is at least $\frac{1}{3}$ but less than $2$ .

Givens

$\frac{4}{a}$ and $\frac{6}{a}$ are integers.
$\frac{1}{3} \le \frac{|b|}{a} < 2$
$a$ and $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

$a \ne 0$ because it is a denominator.
Each allowed value of $|b|$ (other than $0$ ) gives two values of $b$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Condition (a) leaves only a few choices for $a$ , so split into one case per $a$ . In each case, condition (b) becomes a range for $|b|$ , and each value of $|b|$ gives two values of $b$ , one positive and one negative.

3 · Execute5 carry out the plan

1Find the sign of a

#7 Identify Subproblems 7.NS.A.2

In (b) the ratio

\frac{|b|}{a}

is at least

\frac{1}{3}

, so it is positive. Its top

|b|

is never negative, so the bottom

a

must be positive (and

b \ne 0

).

\frac{|b|}{a} \ge \frac{1}{3} > 0,\ |b| \ge 0 \Rightarrow a > 0

A quotient is positive only when top and bottom have the same sign.

2List the choices for a

#7 Identify Subproblems 6.NS.B.4

By (a),

a

is a positive common divisor of

4

and

6

, that is, a divisor of

2

.

a \in \{1, 2\}

Only two cases:

a = 1

,

2

.

3Turn (b) into a range for |b|

#2 Make a Systematic List 6.NS.C.7

Multiply each part of (b) by the positive number

a

.

\frac{a}{3} \le |b| < 2a

Multiplying by a positive number keeps the inequality directions.

4Read off |b| in each case

#2 Make a Systematic List 6.NS.C.7

a = 1

:

\frac{1}{3} \le |b| < 2

.

a = 2

:

\frac{2}{3} \le |b| < 4

.

a=1:\ |b| = 1;\quad a=2:\ |b| = 1, 2, 3

That is

1

and

3

values of

|b|

.

5Count the pairs

#2 Make a Systematic List 6.NS.C.7

Each value of

|b|

gives

b

and

-b

, so double each count and add.

2 \times 1 + 2 \times 3 = 2 + 6 = 8

2 + 6 = 8

pairs.

Answer: 8

4 · Reviewdoes it hold up?

Spot-check a pair from each case: $(1, -1)$ gives $\frac{1}{1} = 1$ ; $(2, 3)$ gives $\frac{3}{2}$ , all in $[\frac{1}{3}, 2)$ . And $(2, 4)$ gives exactly $2$ , correctly excluded.

Another way: Count the positive $b$ first ( $1 + 3 = 4$ pairs) and then double for the negative $b$ : $4 \times 2 = 8$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the common divisors of $4$ and $6$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for quotients to see that $a$ must be positive.
6.NS.C.7 Order and absolute value of rational numbers — Turning the bound on $|b|$ into the integers $b$ at those distances from $0$.

💡Takeaway. Pin down the few possible denominators first, then each case is a simple count.

Variant 2 easy

How many ordered pairs $(a, b)$ of integers satisfy both of the following conditions?

(a) $\frac{6}{a}$ and $\frac{9}{a}$ are both integers.
(b) $\frac{1}{2}\le\frac{|b|}{a}<2$

Your answer

Show solution

1 · Understandwhat's really being asked

Count integer pairs $(a, b)$ where $a$ divides both $6$ and $9$ , and $\frac{|b|}{a}$ is at least $\frac{1}{2}$ but less than $2$ .

Givens

$\frac{6}{a}$ and $\frac{9}{a}$ are integers.
$\frac{1}{2} \le \frac{|b|}{a} < 2$
$a$ and $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

$a \ne 0$ because it is a denominator.
Each allowed value of $|b|$ (other than $0$ ) gives two values of $b$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Condition (a) leaves only a few choices for $a$ , so split into one case per $a$ . In each case, condition (b) becomes a range for $|b|$ , and each value of $|b|$ gives two values of $b$ , one positive and one negative.

3 · Execute5 carry out the plan

1Find the sign of a

#7 Identify Subproblems 7.NS.A.2

In (b) the ratio

\frac{|b|}{a}

is at least

\frac{1}{2}

, so it is positive. Its top

|b|

is never negative, so the bottom

a

must be positive (and

b \ne 0

).

\frac{|b|}{a} \ge \frac{1}{2} > 0,\ |b| \ge 0 \Rightarrow a > 0

A quotient is positive only when top and bottom have the same sign.

2List the choices for a

#7 Identify Subproblems 6.NS.B.4

By (a),

a

is a positive common divisor of

6

and

9

, that is, a divisor of

3

.

a \in \{1, 3\}

Only two cases:

a = 1

,

3

.

3Turn (b) into a range for |b|

#2 Make a Systematic List 6.NS.C.7

Multiply each part of (b) by the positive number

a

.

\frac{a}{2} \le |b| < 2a

Multiplying by a positive number keeps the inequality directions.

4Read off |b| in each case

#2 Make a Systematic List 6.NS.C.7

a = 1

:

\frac{1}{2} \le |b| < 2

.

a = 3

:

1\frac{1}{2} \le |b| < 6

.

a=1:\ |b| = 1;\quad a=3:\ |b| = 2, 3, \dots, 5

That is

1

and

4

values of

|b|

.

5Count the pairs

#2 Make a Systematic List 6.NS.C.7

Each value of

|b|

gives

b

and

-b

, so double each count and add.

2 \times 1 + 2 \times 4 = 2 + 8 = 10

2 + 8 = 10

pairs.

Answer: 10

4 · Reviewdoes it hold up?

Spot-check a pair from each case: $(1, -1)$ gives $\frac{1}{1} = 1$ ; $(3, 5)$ gives $\frac{5}{3}$ , all in $[\frac{1}{2}, 2)$ . And $(3, 6)$ gives exactly $2$ , correctly excluded.

Another way: Count the positive $b$ first ( $1 + 4 = 5$ pairs) and then double for the negative $b$ : $5 \times 2 = 10$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the common divisors of $6$ and $9$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for quotients to see that $a$ must be positive.
6.NS.C.7 Order and absolute value of rational numbers — Turning the bound on $|b|$ into the integers $b$ at those distances from $0$.

💡Takeaway. Pin down the few possible denominators first, then each case is a simple count.

Variant 3 easy

How many ordered pairs $(a, b)$ of integers satisfy both of the following conditions?

(a) $\frac{6}{a}$ and $\frac{10}{a}$ are both integers.
(b) $\frac{1}{2}\le\frac{|b|}{a}<3$

Your answer

Show solution

1 · Understandwhat's really being asked

Count integer pairs $(a, b)$ where $a$ divides both $6$ and $10$ , and $\frac{|b|}{a}$ is at least $\frac{1}{2}$ but less than $3$ .

Givens

$\frac{6}{a}$ and $\frac{10}{a}$ are integers.
$\frac{1}{2} \le \frac{|b|}{a} < 3$
$a$ and $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

$a \ne 0$ because it is a denominator.
Each allowed value of $|b|$ (other than $0$ ) gives two values of $b$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Condition (a) leaves only a few choices for $a$ , so split into one case per $a$ . In each case, condition (b) becomes a range for $|b|$ , and each value of $|b|$ gives two values of $b$ , one positive and one negative.

3 · Execute5 carry out the plan

1Find the sign of a

#7 Identify Subproblems 7.NS.A.2

In (b) the ratio

\frac{|b|}{a}

is at least

\frac{1}{2}

, so it is positive. Its top

|b|

is never negative, so the bottom

a

must be positive (and

b \ne 0

).

\frac{|b|}{a} \ge \frac{1}{2} > 0,\ |b| \ge 0 \Rightarrow a > 0

A quotient is positive only when top and bottom have the same sign.

2List the choices for a

#7 Identify Subproblems 6.NS.B.4

By (a),

a

is a positive common divisor of

6

and

10

, that is, a divisor of

2

.

a \in \{1, 2\}

Only two cases:

a = 1

,

2

.

3Turn (b) into a range for |b|

#2 Make a Systematic List 6.NS.C.7

Multiply each part of (b) by the positive number

a

.

\frac{a}{2} \le |b| < 3a

Multiplying by a positive number keeps the inequality directions.

4Read off |b| in each case

#2 Make a Systematic List 6.NS.C.7

a = 1

:

\frac{1}{2} \le |b| < 3

.

a = 2

:

1 \le |b| < 6

.

a=1:\ |b| = 1, 2;\quad a=2:\ |b| = 1, 2, \dots, 5

That is

2

and

5

values of

|b|

.

5Count the pairs

#2 Make a Systematic List 6.NS.C.7

Each value of

|b|

gives

b

and

-b

, so double each count and add.

2 \times 2 + 2 \times 5 = 4 + 10 = 14

4 + 10 = 14

pairs.

Answer: 14

4 · Reviewdoes it hold up?

Spot-check a pair from each case: $(1, -2)$ gives $\frac{2}{1} = 2$ ; $(2, 5)$ gives $\frac{5}{2}$ , all in $[\frac{1}{2}, 3)$ . And $(2, 6)$ gives exactly $3$ , correctly excluded.

Another way: Count the positive $b$ first ( $2 + 5 = 7$ pairs) and then double for the negative $b$ : $7 \times 2 = 14$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the common divisors of $6$ and $10$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for quotients to see that $a$ must be positive.
6.NS.C.7 Order and absolute value of rational numbers — Turning the bound on $|b|$ into the integers $b$ at those distances from $0$.

💡Takeaway. Pin down the few possible denominators first, then each case is a simple count.

Variant 4 medium

How many ordered pairs $(a, b)$ of integers satisfy both of the following conditions?

(a) $\frac{8}{a}$ and $\frac{12}{a}$ are both integers.
(b) $\frac{1}{3}\le\frac{|b|}{a}<3$

Your answer

Show solution

1 · Understandwhat's really being asked

Count integer pairs $(a, b)$ where $a$ divides both $8$ and $12$ , and $\frac{|b|}{a}$ is at least $\frac{1}{3}$ but less than $3$ .

Givens

$\frac{8}{a}$ and $\frac{12}{a}$ are integers.
$\frac{1}{3} \le \frac{|b|}{a} < 3$
$a$ and $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

$a \ne 0$ because it is a denominator.
Each allowed value of $|b|$ (other than $0$ ) gives two values of $b$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Condition (a) leaves only a few choices for $a$ , so split into one case per $a$ . In each case, condition (b) becomes a range for $|b|$ , and each value of $|b|$ gives two values of $b$ , one positive and one negative.

3 · Execute5 carry out the plan

1Find the sign of a

#7 Identify Subproblems 7.NS.A.2

In (b) the ratio

\frac{|b|}{a}

is at least

\frac{1}{3}

, so it is positive. Its top

|b|

is never negative, so the bottom

a

must be positive (and

b \ne 0

).

\frac{|b|}{a} \ge \frac{1}{3} > 0,\ |b| \ge 0 \Rightarrow a > 0

A quotient is positive only when top and bottom have the same sign.

2List the choices for a

#7 Identify Subproblems 6.NS.B.4

By (a),

a

is a positive common divisor of

8

and

12

, that is, a divisor of

4

.

a \in \{1, 2, 4\}

Only three cases:

a = 1

,

2

,

4

.

3Turn (b) into a range for |b|

#2 Make a Systematic List 6.NS.C.7

Multiply each part of (b) by the positive number

a

.

\frac{a}{3} \le |b| < 3a

Multiplying by a positive number keeps the inequality directions.

4Read off |b| in each case

#2 Make a Systematic List 6.NS.C.7

a = 1

:

\frac{1}{3} \le |b| < 3

.

a = 2

:

\frac{2}{3} \le |b| < 6

.

a = 4

:

1\frac{1}{3} \le |b| < 12

.

a=1:\ |b| = 1, 2;\quad a=2:\ |b| = 1, 2, \dots, 5;\quad a=4:\ |b| = 2, 3, \dots, 11

That is

2

,

5

and

10

values of

|b|

.

5Count the pairs

#2 Make a Systematic List 6.NS.C.7

Each value of

|b|

gives

b

and

-b

, so double each count and add.

2 \times 2 + 2 \times 5 + 2 \times 10 = 4 + 10 + 20 = 34

4 + 10 + 20 = 34

pairs.

Answer: 34

4 · Reviewdoes it hold up?

Spot-check a pair from each case: $(1, -2)$ gives $\frac{2}{1} = 2$ ; $(2, 5)$ gives $\frac{5}{2}$ ; $(4, -11)$ gives $\frac{11}{4}$ , all in $[\frac{1}{3}, 3)$ . And $(4, 12)$ gives exactly $3$ , correctly excluded.

Another way: Count the positive $b$ first ( $2 + 5 + 10 = 17$ pairs) and then double for the negative $b$ : $17 \times 2 = 34$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the common divisors of $8$ and $12$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for quotients to see that $a$ must be positive.
6.NS.C.7 Order and absolute value of rational numbers — Turning the bound on $|b|$ into the integers $b$ at those distances from $0$.

💡Takeaway. Pin down the few possible denominators first, then each case is a simple count.

Variant 5 easy

How many ordered pairs $(a, b)$ of integers satisfy both of the following conditions?

(a) $\frac{8}{a}$ and $\frac{12}{a}$ are both integers.
(b) $\frac{1}{2}\le\frac{|b|}{a}<2$

Your answer

Show solution

1 · Understandwhat's really being asked

Count integer pairs $(a, b)$ where $a$ divides both $8$ and $12$ , and $\frac{|b|}{a}$ is at least $\frac{1}{2}$ but less than $2$ .

Givens

$\frac{8}{a}$ and $\frac{12}{a}$ are integers.
$\frac{1}{2} \le \frac{|b|}{a} < 2$
$a$ and $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

$a \ne 0$ because it is a denominator.
Each allowed value of $|b|$ (other than $0$ ) gives two values of $b$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Condition (a) leaves only a few choices for $a$ , so split into one case per $a$ . In each case, condition (b) becomes a range for $|b|$ , and each value of $|b|$ gives two values of $b$ , one positive and one negative.

3 · Execute5 carry out the plan

1Find the sign of a

#7 Identify Subproblems 7.NS.A.2

In (b) the ratio

\frac{|b|}{a}

is at least

\frac{1}{2}

, so it is positive. Its top

|b|

is never negative, so the bottom

a

must be positive (and

b \ne 0

).

\frac{|b|}{a} \ge \frac{1}{2} > 0,\ |b| \ge 0 \Rightarrow a > 0

A quotient is positive only when top and bottom have the same sign.

2List the choices for a

#7 Identify Subproblems 6.NS.B.4

By (a),

a

is a positive common divisor of

8

and

12

, that is, a divisor of

4

.

a \in \{1, 2, 4\}

Only three cases:

a = 1

,

2

,

4

.

3Turn (b) into a range for |b|

#2 Make a Systematic List 6.NS.C.7

Multiply each part of (b) by the positive number

a

.

\frac{a}{2} \le |b| < 2a

Multiplying by a positive number keeps the inequality directions.

4Read off |b| in each case

#2 Make a Systematic List 6.NS.C.7

a = 1

:

\frac{1}{2} \le |b| < 2

.

a = 2

:

1 \le |b| < 4

.

a = 4

:

2 \le |b| < 8

.

a=1:\ |b| = 1;\quad a=2:\ |b| = 1, 2, 3;\quad a=4:\ |b| = 2, 3, \dots, 7

That is

1

,

3

and

6

values of

|b|

.

5Count the pairs

#2 Make a Systematic List 6.NS.C.7

Each value of

|b|

gives

b

and

-b

, so double each count and add.

2 \times 1 + 2 \times 3 + 2 \times 6 = 2 + 6 + 12 = 20

2 + 6 + 12 = 20

pairs.

Answer: 20

4 · Reviewdoes it hold up?

Spot-check a pair from each case: $(1, -1)$ gives $\frac{1}{1} = 1$ ; $(2, 3)$ gives $\frac{3}{2}$ ; $(4, -7)$ gives $\frac{7}{4}$ , all in $[\frac{1}{2}, 2)$ . And $(4, 8)$ gives exactly $2$ , correctly excluded.

Another way: Count the positive $b$ first ( $1 + 3 + 6 = 10$ pairs) and then double for the negative $b$ : $10 \times 2 = 20$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the common divisors of $8$ and $12$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for quotients to see that $a$ must be positive.
6.NS.C.7 Order and absolute value of rational numbers — Turning the bound on $|b|$ into the integers $b$ at those distances from $0$.

💡Takeaway. Pin down the few possible denominators first, then each case is a simple count.

Variant 6 medium

How many ordered pairs $(a, b)$ of integers satisfy both of the following conditions?

(a) $\frac{9}{a}$ and $\frac{15}{a}$ are both integers.
(b) $\frac{3}{4}\le\frac{|b|}{a}<2$

Your answer

Show solution

1 · Understandwhat's really being asked

Count integer pairs $(a, b)$ where $a$ divides both $9$ and $15$ , and $\frac{|b|}{a}$ is at least $\frac{3}{4}$ but less than $2$ .

Givens

$\frac{9}{a}$ and $\frac{15}{a}$ are integers.
$\frac{3}{4} \le \frac{|b|}{a} < 2$
$a$ and $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

$a \ne 0$ because it is a denominator.
Each allowed value of $|b|$ (other than $0$ ) gives two values of $b$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Condition (a) leaves only a few choices for $a$ , so split into one case per $a$ . In each case, condition (b) becomes a range for $|b|$ , and each value of $|b|$ gives two values of $b$ , one positive and one negative.

3 · Execute5 carry out the plan

1Find the sign of a

#7 Identify Subproblems 7.NS.A.2

In (b) the ratio

\frac{|b|}{a}

is at least

\frac{3}{4}

, so it is positive. Its top

|b|

is never negative, so the bottom

a

must be positive (and

b \ne 0

).

\frac{|b|}{a} \ge \frac{3}{4} > 0,\ |b| \ge 0 \Rightarrow a > 0

A quotient is positive only when top and bottom have the same sign.

2List the choices for a

#7 Identify Subproblems 6.NS.B.4

By (a),

a

is a positive common divisor of

9

and

15

, that is, a divisor of

3

.

a \in \{1, 3\}

Only two cases:

a = 1

,

3

.

3Turn (b) into a range for |b|

#2 Make a Systematic List 6.NS.C.7

Multiply each part of (b) by the positive number

a

.

\frac{3a}{4} \le |b| < 2a

Multiplying by a positive number keeps the inequality directions.

4Read off |b| in each case

#2 Make a Systematic List 6.NS.C.7

a = 1

:

\frac{3}{4} \le |b| < 2

.

a = 3

:

2\frac{1}{4} \le |b| < 6

.

a=1:\ |b| = 1;\quad a=3:\ |b| = 3, 4, 5

That is

1

and

3

values of

|b|

.

5Count the pairs

#2 Make a Systematic List 6.NS.C.7

Each value of

|b|

gives

b

and

-b

, so double each count and add.

2 \times 1 + 2 \times 3 = 2 + 6 = 8

2 + 6 = 8

pairs.

Answer: 8

4 · Reviewdoes it hold up?

Spot-check a pair from each case: $(1, -1)$ gives $\frac{1}{1} = 1$ ; $(3, 5)$ gives $\frac{5}{3}$ , all in $[\frac{3}{4}, 2)$ . And $(3, 6)$ gives exactly $2$ , correctly excluded.

Another way: Count the positive $b$ first ( $1 + 3 = 4$ pairs) and then double for the negative $b$ : $4 \times 2 = 8$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the common divisors of $9$ and $15$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for quotients to see that $a$ must be positive.
6.NS.C.7 Order and absolute value of rational numbers — Turning the bound on $|b|$ into the integers $b$ at those distances from $0$.

💡Takeaway. Pin down the few possible denominators first, then each case is a simple count.

Variant 7 medium

How many ordered pairs $(a, b)$ of integers satisfy both of the following conditions?

(a) $\frac{10}{a}$ and $\frac{15}{a}$ are both integers.
(b) $\frac{1}{3}\le\frac{|b|}{a}<2$

Your answer

Show solution

1 · Understandwhat's really being asked

Count integer pairs $(a, b)$ where $a$ divides both $10$ and $15$ , and $\frac{|b|}{a}$ is at least $\frac{1}{3}$ but less than $2$ .

Givens

$\frac{10}{a}$ and $\frac{15}{a}$ are integers.
$\frac{1}{3} \le \frac{|b|}{a} < 2$
$a$ and $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

$a \ne 0$ because it is a denominator.
Each allowed value of $|b|$ (other than $0$ ) gives two values of $b$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Condition (a) leaves only a few choices for $a$ , so split into one case per $a$ . In each case, condition (b) becomes a range for $|b|$ , and each value of $|b|$ gives two values of $b$ , one positive and one negative.

3 · Execute5 carry out the plan

1Find the sign of a

#7 Identify Subproblems 7.NS.A.2

In (b) the ratio

\frac{|b|}{a}

is at least

\frac{1}{3}

, so it is positive. Its top

|b|

is never negative, so the bottom

a

must be positive (and

b \ne 0

).

\frac{|b|}{a} \ge \frac{1}{3} > 0,\ |b| \ge 0 \Rightarrow a > 0

A quotient is positive only when top and bottom have the same sign.

2List the choices for a

#7 Identify Subproblems 6.NS.B.4

By (a),

a

is a positive common divisor of

10

and

15

, that is, a divisor of

5

.

a \in \{1, 5\}

Only two cases:

a = 1

,

5

.

3Turn (b) into a range for |b|

#2 Make a Systematic List 6.NS.C.7

Multiply each part of (b) by the positive number

a

.

\frac{a}{3} \le |b| < 2a

Multiplying by a positive number keeps the inequality directions.

4Read off |b| in each case

#2 Make a Systematic List 6.NS.C.7

a = 1

:

\frac{1}{3} \le |b| < 2

.

a = 5

:

1\frac{2}{3} \le |b| < 10

.

a=1:\ |b| = 1;\quad a=5:\ |b| = 2, 3, \dots, 9

That is

1

and

8

values of

|b|

.

5Count the pairs

#2 Make a Systematic List 6.NS.C.7

Each value of

|b|

gives

b

and

-b

, so double each count and add.

2 \times 1 + 2 \times 8 = 2 + 16 = 18

2 + 16 = 18

pairs.

Answer: 18

4 · Reviewdoes it hold up?

Spot-check a pair from each case: $(1, -1)$ gives $\frac{1}{1} = 1$ ; $(5, 9)$ gives $\frac{9}{5}$ , all in $[\frac{1}{3}, 2)$ . And $(5, 10)$ gives exactly $2$ , correctly excluded.

Another way: Count the positive $b$ first ( $1 + 8 = 9$ pairs) and then double for the negative $b$ : $9 \times 2 = 18$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the common divisors of $10$ and $15$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for quotients to see that $a$ must be positive.
6.NS.C.7 Order and absolute value of rational numbers — Turning the bound on $|b|$ into the integers $b$ at those distances from $0$.

💡Takeaway. Pin down the few possible denominators first, then each case is a simple count.

Variant 8 medium

How many ordered pairs $(a, b)$ of integers satisfy both of the following conditions?

(a) $\frac{12}{a}$ and $\frac{18}{a}$ are both integers.
(b) $\frac{1}{2}\le\frac{|b|}{a}<2$

Your answer

Show solution

1 · Understandwhat's really being asked

Count integer pairs $(a, b)$ where $a$ divides both $12$ and $18$ , and $\frac{|b|}{a}$ is at least $\frac{1}{2}$ but less than $2$ .

Givens

$\frac{12}{a}$ and $\frac{18}{a}$ are integers.
$\frac{1}{2} \le \frac{|b|}{a} < 2$
$a$ and $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

$a \ne 0$ because it is a denominator.
Each allowed value of $|b|$ (other than $0$ ) gives two values of $b$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Condition (a) leaves only a few choices for $a$ , so split into one case per $a$ . In each case, condition (b) becomes a range for $|b|$ , and each value of $|b|$ gives two values of $b$ , one positive and one negative.

3 · Execute5 carry out the plan

1Find the sign of a

#7 Identify Subproblems 7.NS.A.2

In (b) the ratio

\frac{|b|}{a}

is at least

\frac{1}{2}

, so it is positive. Its top

|b|

is never negative, so the bottom

a

must be positive (and

b \ne 0

).

\frac{|b|}{a} \ge \frac{1}{2} > 0,\ |b| \ge 0 \Rightarrow a > 0

A quotient is positive only when top and bottom have the same sign.

2List the choices for a

#7 Identify Subproblems 6.NS.B.4

By (a),

a

is a positive common divisor of

12

and

18

, that is, a divisor of

6

.

a \in \{1, 2, 3, 6\}

Only four cases:

a = 1

,

2

,

3

,

6

.

3Turn (b) into a range for |b|

#2 Make a Systematic List 6.NS.C.7

Multiply each part of (b) by the positive number

a

.

\frac{a}{2} \le |b| < 2a

Multiplying by a positive number keeps the inequality directions.

4Read off |b| in each case

#2 Make a Systematic List 6.NS.C.7

a = 1

:

\frac{1}{2} \le |b| < 2

.

a = 2

:

1 \le |b| < 4

.

a = 3

:

1\frac{1}{2} \le |b| < 6

.

a = 6

:

3 \le |b| < 12

.

a=1:\ |b| = 1;\quad a=2:\ |b| = 1, 2, 3;\quad a=3:\ |b| = 2, 3, \dots, 5;\quad a=6:\ |b| = 3, 4, \dots, 11

That is

1

,

3

,

4

and

9

values of

|b|

.

5Count the pairs

#2 Make a Systematic List 6.NS.C.7

Each value of

|b|

gives

b

and

-b

, so double each count and add.

2 \times 1 + 2 \times 3 + 2 \times 4 + 2 \times 9 = 2 + 6 + 8 + 18 = 34

2 + 6 + 8 + 18 = 34

pairs.

Answer: 34

4 · Reviewdoes it hold up?

Spot-check a pair from each case: $(1, -1)$ gives $\frac{1}{1} = 1$ ; $(2, 3)$ gives $\frac{3}{2}$ ; $(3, -5)$ gives $\frac{5}{3}$ ; $(6, 11)$ gives $\frac{11}{6}$ , all in $[\frac{1}{2}, 2)$ . And $(6, 12)$ gives exactly $2$ , correctly excluded.

Another way: Count the positive $b$ first ( $1 + 3 + 4 + 9 = 17$ pairs) and then double for the negative $b$ : $17 \times 2 = 34$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the common divisors of $12$ and $18$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for quotients to see that $a$ must be positive.
6.NS.C.7 Order and absolute value of rational numbers — Turning the bound on $|b|$ into the integers $b$ at those distances from $0$.

💡Takeaway. Pin down the few possible denominators first, then each case is a simple count.

Variant 9 hard

How many ordered pairs $(a, b)$ of integers satisfy both of the following conditions?

(a) $\frac{12}{a}$ and $\frac{20}{a}$ are both integers.
(b) $\frac{2}{3}\le\frac{|b|}{a}<2$

Your answer

Show solution

1 · Understandwhat's really being asked

Count integer pairs $(a, b)$ where $a$ divides both $12$ and $20$ , and $\frac{|b|}{a}$ is at least $\frac{2}{3}$ but less than $2$ .

Givens

$\frac{12}{a}$ and $\frac{20}{a}$ are integers.
$\frac{2}{3} \le \frac{|b|}{a} < 2$
$a$ and $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

$a \ne 0$ because it is a denominator.
Each allowed value of $|b|$ (other than $0$ ) gives two values of $b$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Condition (a) leaves only a few choices for $a$ , so split into one case per $a$ . In each case, condition (b) becomes a range for $|b|$ , and each value of $|b|$ gives two values of $b$ , one positive and one negative.

3 · Execute5 carry out the plan

1Find the sign of a

#7 Identify Subproblems 7.NS.A.2

In (b) the ratio

\frac{|b|}{a}

is at least

\frac{2}{3}

, so it is positive. Its top

|b|

is never negative, so the bottom

a

must be positive (and

b \ne 0

).

\frac{|b|}{a} \ge \frac{2}{3} > 0,\ |b| \ge 0 \Rightarrow a > 0

A quotient is positive only when top and bottom have the same sign.

2List the choices for a

#7 Identify Subproblems 6.NS.B.4

By (a),

a

is a positive common divisor of

12

and

20

, that is, a divisor of

4

.

a \in \{1, 2, 4\}

Only three cases:

a = 1

,

2

,

4

.

3Turn (b) into a range for |b|

#2 Make a Systematic List 6.NS.C.7

Multiply each part of (b) by the positive number

a

.

\frac{2a}{3} \le |b| < 2a

Multiplying by a positive number keeps the inequality directions.

4Read off |b| in each case

#2 Make a Systematic List 6.NS.C.7

a = 1

:

\frac{2}{3} \le |b| < 2

.

a = 2

:

1\frac{1}{3} \le |b| < 4

.

a = 4

:

2\frac{2}{3} \le |b| < 8

.

a=1:\ |b| = 1;\quad a=2:\ |b| = 2, 3;\quad a=4:\ |b| = 3, 4, \dots, 7

That is

1

,

2

and

5

values of

|b|

.

5Count the pairs

#2 Make a Systematic List 6.NS.C.7

Each value of

|b|

gives

b

and

-b

, so double each count and add.

2 \times 1 + 2 \times 2 + 2 \times 5 = 2 + 4 + 10 = 16

2 + 4 + 10 = 16

pairs.

Answer: 16

4 · Reviewdoes it hold up?

Spot-check a pair from each case: $(1, -1)$ gives $\frac{1}{1} = 1$ ; $(2, 3)$ gives $\frac{3}{2}$ ; $(4, -7)$ gives $\frac{7}{4}$ , all in $[\frac{2}{3}, 2)$ . And $(4, 8)$ gives exactly $2$ , correctly excluded.

Another way: Count the positive $b$ first ( $1 + 2 + 5 = 8$ pairs) and then double for the negative $b$ : $8 \times 2 = 16$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the common divisors of $12$ and $20$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for quotients to see that $a$ must be positive.
6.NS.C.7 Order and absolute value of rational numbers — Turning the bound on $|b|$ into the integers $b$ at those distances from $0$.

💡Takeaway. Pin down the few possible denominators first, then each case is a simple count.

Variant 10 hard

How many ordered pairs $(a, b)$ of integers satisfy both of the following conditions?

(a) $\frac{14}{a}$ and $\frac{21}{a}$ are both integers.
(b) $\frac{1}{2}\le\frac{|b|}{a}<2$

Your answer

Show solution

1 · Understandwhat's really being asked

Count integer pairs $(a, b)$ where $a$ divides both $14$ and $21$ , and $\frac{|b|}{a}$ is at least $\frac{1}{2}$ but less than $2$ .

Givens

$\frac{14}{a}$ and $\frac{21}{a}$ are integers.
$\frac{1}{2} \le \frac{|b|}{a} < 2$
$a$ and $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

$a \ne 0$ because it is a denominator.
Each allowed value of $|b|$ (other than $0$ ) gives two values of $b$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Condition (a) leaves only a few choices for $a$ , so split into one case per $a$ . In each case, condition (b) becomes a range for $|b|$ , and each value of $|b|$ gives two values of $b$ , one positive and one negative.

3 · Execute5 carry out the plan

1Find the sign of a

#7 Identify Subproblems 7.NS.A.2

In (b) the ratio

\frac{|b|}{a}

is at least

\frac{1}{2}

, so it is positive. Its top

|b|

is never negative, so the bottom

a

must be positive (and

b \ne 0

).

\frac{|b|}{a} \ge \frac{1}{2} > 0,\ |b| \ge 0 \Rightarrow a > 0

A quotient is positive only when top and bottom have the same sign.

2List the choices for a

#7 Identify Subproblems 6.NS.B.4

By (a),

a

is a positive common divisor of

14

and

21

, that is, a divisor of

7

.

a \in \{1, 7\}

Only two cases:

a = 1

,

7

.

3Turn (b) into a range for |b|

#2 Make a Systematic List 6.NS.C.7

Multiply each part of (b) by the positive number

a

.

\frac{a}{2} \le |b| < 2a

Multiplying by a positive number keeps the inequality directions.

4Read off |b| in each case

#2 Make a Systematic List 6.NS.C.7

a = 1

:

\frac{1}{2} \le |b| < 2

.

a = 7

:

3\frac{1}{2} \le |b| < 14

.

a=1:\ |b| = 1;\quad a=7:\ |b| = 4, 5, \dots, 13

That is

1

and

10

values of

|b|

.

5Count the pairs

#2 Make a Systematic List 6.NS.C.7

Each value of

|b|

gives

b

and

-b

, so double each count and add.

2 \times 1 + 2 \times 10 = 2 + 20 = 22

2 + 20 = 22

pairs.

Answer: 22

4 · Reviewdoes it hold up?

Spot-check a pair from each case: $(1, -1)$ gives $\frac{1}{1} = 1$ ; $(7, 13)$ gives $\frac{13}{7}$ , all in $[\frac{1}{2}, 2)$ . And $(7, 14)$ gives exactly $2$ , correctly excluded.

Another way: Count the positive $b$ first ( $1 + 10 = 11$ pairs) and then double for the negative $b$ : $11 \times 2 = 22$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the common divisors of $14$ and $21$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for quotients to see that $a$ must be positive.
6.NS.C.7 Order and absolute value of rational numbers — Turning the bound on $|b|$ into the integers $b$ at those distances from $0$.

💡Takeaway. Pin down the few possible denominators first, then each case is a simple count.

Variant 11 hard

How many ordered pairs $(a, b)$ of integers satisfy both of the following conditions?

(a) $\frac{18}{a}$ and $\frac{27}{a}$ are both integers.
(b) $\frac{1}{2}\le\frac{|b|}{a}<2$

Your answer

Show solution

1 · Understandwhat's really being asked

Count integer pairs $(a, b)$ where $a$ divides both $18$ and $27$ , and $\frac{|b|}{a}$ is at least $\frac{1}{2}$ but less than $2$ .

Givens

$\frac{18}{a}$ and $\frac{27}{a}$ are integers.
$\frac{1}{2} \le \frac{|b|}{a} < 2$
$a$ and $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

$a \ne 0$ because it is a denominator.
Each allowed value of $|b|$ (other than $0$ ) gives two values of $b$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Condition (a) leaves only a few choices for $a$ , so split into one case per $a$ . In each case, condition (b) becomes a range for $|b|$ , and each value of $|b|$ gives two values of $b$ , one positive and one negative.

3 · Execute5 carry out the plan

1Find the sign of a

#7 Identify Subproblems 7.NS.A.2

In (b) the ratio

\frac{|b|}{a}

is at least

\frac{1}{2}

, so it is positive. Its top

|b|

is never negative, so the bottom

a

must be positive (and

b \ne 0

).

\frac{|b|}{a} \ge \frac{1}{2} > 0,\ |b| \ge 0 \Rightarrow a > 0

A quotient is positive only when top and bottom have the same sign.

2List the choices for a

#7 Identify Subproblems 6.NS.B.4

By (a),

a

is a positive common divisor of

18

and

27

, that is, a divisor of

9

.

a \in \{1, 3, 9\}

Only three cases:

a = 1

,

3

,

9

.

3Turn (b) into a range for |b|

#2 Make a Systematic List 6.NS.C.7

Multiply each part of (b) by the positive number

a

.

\frac{a}{2} \le |b| < 2a

Multiplying by a positive number keeps the inequality directions.

4Read off |b| in each case

#2 Make a Systematic List 6.NS.C.7

a = 1

:

\frac{1}{2} \le |b| < 2

.

a = 3

:

1\frac{1}{2} \le |b| < 6

.

a = 9

:

4\frac{1}{2} \le |b| < 18

.

a=1:\ |b| = 1;\quad a=3:\ |b| = 2, 3, \dots, 5;\quad a=9:\ |b| = 5, 6, \dots, 17

That is

1

,

4

and

13

values of

|b|

.

5Count the pairs

#2 Make a Systematic List 6.NS.C.7

Each value of

|b|

gives

b

and

-b

, so double each count and add.

2 \times 1 + 2 \times 4 + 2 \times 13 = 2 + 8 + 26 = 36

2 + 8 + 26 = 36

pairs.

Answer: 36

4 · Reviewdoes it hold up?

Spot-check a pair from each case: $(1, -1)$ gives $\frac{1}{1} = 1$ ; $(3, 5)$ gives $\frac{5}{3}$ ; $(9, -17)$ gives $\frac{17}{9}$ , all in $[\frac{1}{2}, 2)$ . And $(9, 18)$ gives exactly $2$ , correctly excluded.

Another way: Count the positive $b$ first ( $1 + 4 + 13 = 18$ pairs) and then double for the negative $b$ : $18 \times 2 = 36$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the common divisors of $18$ and $27$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for quotients to see that $a$ must be positive.
6.NS.C.7 Order and absolute value of rational numbers — Turning the bound on $|b|$ into the integers $b$ at those distances from $0$.

💡Takeaway. Pin down the few possible denominators first, then each case is a simple count.

Variant 12 hard

How many ordered pairs $(a, b)$ of integers satisfy both of the following conditions?

(a) $\frac{12}{a}$ and $\frac{30}{a}$ are both integers.
(b) $\frac{2}{3}\le\frac{|b|}{a}<2$

Your answer

Show solution

1 · Understandwhat's really being asked

Count integer pairs $(a, b)$ where $a$ divides both $12$ and $30$ , and $\frac{|b|}{a}$ is at least $\frac{2}{3}$ but less than $2$ .

Givens

$\frac{12}{a}$ and $\frac{30}{a}$ are integers.
$\frac{2}{3} \le \frac{|b|}{a} < 2$
$a$ and $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

$a \ne 0$ because it is a denominator.
Each allowed value of $|b|$ (other than $0$ ) gives two values of $b$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #7 Identify Subproblems

Condition (a) leaves only a few choices for $a$ , so split into one case per $a$ . In each case, condition (b) becomes a range for $|b|$ , and each value of $|b|$ gives two values of $b$ , one positive and one negative.

3 · Execute5 carry out the plan

1Find the sign of a

#7 Identify Subproblems 7.NS.A.2

In (b) the ratio

\frac{|b|}{a}

is at least

\frac{2}{3}

, so it is positive. Its top

|b|

is never negative, so the bottom

a

must be positive (and

b \ne 0

).

\frac{|b|}{a} \ge \frac{2}{3} > 0,\ |b| \ge 0 \Rightarrow a > 0

A quotient is positive only when top and bottom have the same sign.

2List the choices for a

#7 Identify Subproblems 6.NS.B.4

By (a),

a

is a positive common divisor of

12

and

30

, that is, a divisor of

6

.

a \in \{1, 2, 3, 6\}

Only four cases:

a = 1

,

2

,

3

,

6

.

3Turn (b) into a range for |b|

#2 Make a Systematic List 6.NS.C.7

Multiply each part of (b) by the positive number

a

.

\frac{2a}{3} \le |b| < 2a

Multiplying by a positive number keeps the inequality directions.

4Read off |b| in each case

#2 Make a Systematic List 6.NS.C.7

a = 1

:

\frac{2}{3} \le |b| < 2

.

a = 2

:

1\frac{1}{3} \le |b| < 4

.

a = 3

:

2 \le |b| < 6

.

a = 6

:

4 \le |b| < 12

.

a=1:\ |b| = 1;\quad a=2:\ |b| = 2, 3;\quad a=3:\ |b| = 2, 3, \dots, 5;\quad a=6:\ |b| = 4, 5, \dots, 11

That is

1

,

2

,

4

and

8

values of

|b|

.

5Count the pairs

#2 Make a Systematic List 6.NS.C.7

Each value of

|b|

gives

b

and

-b

, so double each count and add.

2 \times 1 + 2 \times 2 + 2 \times 4 + 2 \times 8 = 2 + 4 + 8 + 16 = 30

2 + 4 + 8 + 16 = 30

pairs.

Answer: 30

4 · Reviewdoes it hold up?

Spot-check a pair from each case: $(1, -1)$ gives $\frac{1}{1} = 1$ ; $(2, 3)$ gives $\frac{3}{2}$ ; $(3, -5)$ gives $\frac{5}{3}$ ; $(6, 11)$ gives $\frac{11}{6}$ , all in $[\frac{2}{3}, 2)$ . And $(6, 12)$ gives exactly $2$ , correctly excluded.

Another way: Count the positive $b$ first ( $1 + 2 + 4 + 8 = 15$ pairs) and then double for the negative $b$ : $15 \times 2 = 30$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the common divisors of $12$ and $30$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Using the sign rule for quotients to see that $a$ must be positive.
6.NS.C.7 Order and absolute value of rational numbers — Turning the bound on $|b|$ into the integers $b$ at those distances from $0$.

💡Takeaway. Pin down the few possible denominators first, then each case is a simple count.

Split by common divisors, count |b| values · 12 practice problems

Generated variants — 12

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Where next?