Practice · Count numerators that cancel the denominator's 3s · Sensim Math

Variant 1 easy

Among the fractions with denominator $36$ that lie between $0.\overline{3}$ and $0.\overline{8}$ , how many can be written as terminating decimals?

Your answer

Show solution

1 · Understandwhat's really being asked

Look at fractions $\frac{x}{36}$ that sit strictly between $0.\overline{3}$ and $0.\overline{8}$ . Count how many of them are terminating decimals.

Givens

The denominator is $36$ .
The fraction lies between $0.\overline{3}$ and $0.\overline{8}$ .

Unknowns

How many of these fractions are terminating decimals.

Constraints

A fraction in lowest terms terminates exactly when its denominator has no prime factor other than $2$ and $5$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#2 Make a Systematic List

The bounds are repeating decimals, so first rewrite them as fractions over $36$ ; that turns the question into a range of whole-number numerators. Then factor $36$ to see which numerators remove the troublesome prime $3$ , and list them.

3 · Execute3 carry out the plan

1Rewrite the bounds over 36

#15 Organize Information in More Ways 8.NS.A.1

A single repeating digit

d

after the point is

\frac{d}{9}

. Scale both bounds up to denominator

36

.

0.\overline{3} = \frac{3}{9} = \frac{12}{36}, \quad 0.\overline{8} = \frac{8}{9} = \frac{32}{36}

So the numerator

x

must satisfy

12 < x < 32

.

2Find which numerators terminate

#7 Identify Subproblems 8.NS.A.1

Factor the denominator:

36 = 2^2 \times 3^2

. The

2

is harmless; the factor

3^2 = 9

must cancel completely, so

x

must be a multiple of

9

.

\frac{x}{36} = \frac{x}{2^2 \times 3^2}

If any

3

is left in the denominator, the decimal repeats forever.

3List the multiples of 9 in range

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

The multiples of

9

strictly between

12

and

32

are

18

and

27

.

9 \times 2 = 18, \quad 9 \times 3 = 27

The next ones,

9

and

36

, fall outside the range, so there are

2

.

Answer: 2

4 · Reviewdoes it hold up?

Check each: $\frac{18}{36} = \frac{1}{2} = 0.5$ , $\frac{27}{36} = \frac{3}{4} = 0.75$ . All terminate and all lie between $0.333\ldots$ and $0.888\ldots$ , while a neighbour such as $\frac{24}{36} = \frac{2}{3} = 0.\overline{6}$ does not terminate.

Another way: Count the complement: there are $19$ numerators from $13$ to $31$ ; every one except the $2$ multiples of $9$ leaves a factor $3$ in the denominator and repeats.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Turning the repeating bounds into fractions and deciding which fractions have terminating decimals.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that the numerator must be a multiple of $9$ and listing those multiples.

💡Takeaway. Put everything over one denominator, then just ask which numerators cancel the 3s.

Variant 2 easy

Among the fractions with denominator $45$ that lie between $0.\overline{2}$ and $0.\overline{8}$ , how many can be written as terminating decimals?

Your answer

Show solution

1 · Understandwhat's really being asked

Look at fractions $\frac{x}{45}$ that sit strictly between $0.\overline{2}$ and $0.\overline{8}$ . Count how many of them are terminating decimals.

Givens

The denominator is $45$ .
The fraction lies between $0.\overline{2}$ and $0.\overline{8}$ .

Unknowns

How many of these fractions are terminating decimals.

Constraints

A fraction in lowest terms terminates exactly when its denominator has no prime factor other than $2$ and $5$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#2 Make a Systematic List

The bounds are repeating decimals, so first rewrite them as fractions over $45$ ; that turns the question into a range of whole-number numerators. Then factor $45$ to see which numerators remove the troublesome prime $3$ , and list them.

3 · Execute3 carry out the plan

1Rewrite the bounds over 45

#15 Organize Information in More Ways 8.NS.A.1

A single repeating digit

d

after the point is

\frac{d}{9}

. Scale both bounds up to denominator

45

.

0.\overline{2} = \frac{2}{9} = \frac{10}{45}, \quad 0.\overline{8} = \frac{8}{9} = \frac{40}{45}

So the numerator

x

must satisfy

10 < x < 40

.

2Find which numerators terminate

#7 Identify Subproblems 8.NS.A.1

Factor the denominator:

45 = 3^2 \times 5

. The

5

is harmless; the factor

3^2 = 9

must cancel completely, so

x

must be a multiple of

9

.

\frac{x}{45} = \frac{x}{3^2 \times 5}

If any

3

is left in the denominator, the decimal repeats forever.

3List the multiples of 9 in range

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

The multiples of

9

strictly between

10

and

40

are

18

,

27

,

36

.

9 \times 2 = 18, \quad 9 \times 3 = 27, \quad 9 \times 4 = 36

The next ones,

9

and

45

, fall outside the range, so there are

3

.

Answer: 3

4 · Reviewdoes it hold up?

Check each: $\frac{18}{45} = \frac{2}{5} = 0.4$ , $\frac{27}{45} = \frac{3}{5} = 0.6$ , $\frac{36}{45} = \frac{4}{5} = 0.8$ . All terminate and all lie between $0.222\ldots$ and $0.888\ldots$ , while a neighbour such as $\frac{15}{45} = \frac{1}{3} = 0.\overline{3}$ does not terminate.

Another way: Count the complement: there are $29$ numerators from $11$ to $39$ ; every one except the $3$ multiples of $9$ leaves a factor $3$ in the denominator and repeats.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Turning the repeating bounds into fractions and deciding which fractions have terminating decimals.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that the numerator must be a multiple of $9$ and listing those multiples.

💡Takeaway. Put everything over one denominator, then just ask which numerators cancel the 3s.

Variant 3 easy

Among the fractions with denominator $72$ that lie between $0.\overline{3}$ and $0.\overline{7}$ , how many can be written as terminating decimals?

Your answer

Show solution

1 · Understandwhat's really being asked

Look at fractions $\frac{x}{72}$ that sit strictly between $0.\overline{3}$ and $0.\overline{7}$ . Count how many of them are terminating decimals.

Givens

The denominator is $72$ .
The fraction lies between $0.\overline{3}$ and $0.\overline{7}$ .

Unknowns

How many of these fractions are terminating decimals.

Constraints

A fraction in lowest terms terminates exactly when its denominator has no prime factor other than $2$ and $5$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#2 Make a Systematic List

The bounds are repeating decimals, so first rewrite them as fractions over $72$ ; that turns the question into a range of whole-number numerators. Then factor $72$ to see which numerators remove the troublesome prime $3$ , and list them.

3 · Execute3 carry out the plan

1Rewrite the bounds over 72

#15 Organize Information in More Ways 8.NS.A.1

A single repeating digit

d

after the point is

\frac{d}{9}

. Scale both bounds up to denominator

72

.

0.\overline{3} = \frac{3}{9} = \frac{24}{72}, \quad 0.\overline{7} = \frac{7}{9} = \frac{56}{72}

So the numerator

x

must satisfy

24 < x < 56

.

2Find which numerators terminate

#7 Identify Subproblems 8.NS.A.1

Factor the denominator:

72 = 2^3 \times 3^2

. The

2

is harmless; the factor

3^2 = 9

must cancel completely, so

x

must be a multiple of

9

.

\frac{x}{72} = \frac{x}{2^3 \times 3^2}

If any

3

is left in the denominator, the decimal repeats forever.

3List the multiples of 9 in range

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

The multiples of

9

strictly between

24

and

56

are

27

,

36

,

45

,

54

.

9 \times 3 = 27, \quad 9 \times 4 = 36, \quad 9 \times 5 = 45, \quad 9 \times 6 = 54

The next ones,

18

and

63

, fall outside the range, so there are

4

.

Answer: 4

4 · Reviewdoes it hold up?

Check each: $\frac{27}{72} = \frac{3}{8} = 0.375$ , $\frac{36}{72} = \frac{1}{2} = 0.5$ , $\frac{45}{72} = \frac{5}{8} = 0.625$ , $\frac{54}{72} = \frac{3}{4} = 0.75$ . All terminate and all lie between $0.333\ldots$ and $0.777\ldots$ , while a neighbour such as $\frac{48}{72} = \frac{2}{3} = 0.\overline{6}$ does not terminate.

Another way: Count the complement: there are $31$ numerators from $25$ to $55$ ; every one except the $4$ multiples of $9$ leaves a factor $3$ in the denominator and repeats.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Turning the repeating bounds into fractions and deciding which fractions have terminating decimals.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that the numerator must be a multiple of $9$ and listing those multiples.

💡Takeaway. Put everything over one denominator, then just ask which numerators cancel the 3s.

Variant 4 easy

Among the fractions with denominator $72$ that lie between $0.\overline{5}$ and $0.\overline{8}$ , how many can be written as terminating decimals?

Your answer

Show solution

1 · Understandwhat's really being asked

Look at fractions $\frac{x}{72}$ that sit strictly between $0.\overline{5}$ and $0.\overline{8}$ . Count how many of them are terminating decimals.

Givens

The denominator is $72$ .
The fraction lies between $0.\overline{5}$ and $0.\overline{8}$ .

Unknowns

How many of these fractions are terminating decimals.

Constraints

A fraction in lowest terms terminates exactly when its denominator has no prime factor other than $2$ and $5$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#2 Make a Systematic List

The bounds are repeating decimals, so first rewrite them as fractions over $72$ ; that turns the question into a range of whole-number numerators. Then factor $72$ to see which numerators remove the troublesome prime $3$ , and list them.

3 · Execute3 carry out the plan

1Rewrite the bounds over 72

#15 Organize Information in More Ways 8.NS.A.1

A single repeating digit

d

after the point is

\frac{d}{9}

. Scale both bounds up to denominator

72

.

0.\overline{5} = \frac{5}{9} = \frac{40}{72}, \quad 0.\overline{8} = \frac{8}{9} = \frac{64}{72}

So the numerator

x

must satisfy

40 < x < 64

.

2Find which numerators terminate

#7 Identify Subproblems 8.NS.A.1

Factor the denominator:

72 = 2^3 \times 3^2

. The

2

is harmless; the factor

3^2 = 9

must cancel completely, so

x

must be a multiple of

9

.

\frac{x}{72} = \frac{x}{2^3 \times 3^2}

If any

3

is left in the denominator, the decimal repeats forever.

3List the multiples of 9 in range

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

The multiples of

9

strictly between

40

and

64

are

45

,

54

,

63

.

9 \times 5 = 45, \quad 9 \times 6 = 54, \quad 9 \times 7 = 63

The next ones,

36

and

72

, fall outside the range, so there are

3

.

Answer: 3

4 · Reviewdoes it hold up?

Check each: $\frac{45}{72} = \frac{5}{8} = 0.625$ , $\frac{54}{72} = \frac{3}{4} = 0.75$ , $\frac{63}{72} = \frac{7}{8} = 0.875$ . All terminate and all lie between $0.555\ldots$ and $0.888\ldots$ , while a neighbour such as $\frac{48}{72} = \frac{2}{3} = 0.\overline{6}$ does not terminate.

Another way: Count the complement: there are $23$ numerators from $41$ to $63$ ; every one except the $3$ multiples of $9$ leaves a factor $3$ in the denominator and repeats.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Turning the repeating bounds into fractions and deciding which fractions have terminating decimals.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that the numerator must be a multiple of $9$ and listing those multiples.

💡Takeaway. Put everything over one denominator, then just ask which numerators cancel the 3s.

Variant 5 medium

Among the fractions with denominator $90$ that lie between $0.\overline{2}$ and $0.\overline{7}$ , how many can be written as terminating decimals?

Your answer

Show solution

1 · Understandwhat's really being asked

Look at fractions $\frac{x}{90}$ that sit strictly between $0.\overline{2}$ and $0.\overline{7}$ . Count how many of them are terminating decimals.

Givens

The denominator is $90$ .
The fraction lies between $0.\overline{2}$ and $0.\overline{7}$ .

Unknowns

How many of these fractions are terminating decimals.

Constraints

A fraction in lowest terms terminates exactly when its denominator has no prime factor other than $2$ and $5$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#2 Make a Systematic List

The bounds are repeating decimals, so first rewrite them as fractions over $90$ ; that turns the question into a range of whole-number numerators. Then factor $90$ to see which numerators remove the troublesome prime $3$ , and list them.

3 · Execute3 carry out the plan

1Rewrite the bounds over 90

#15 Organize Information in More Ways 8.NS.A.1

A single repeating digit

d

after the point is

\frac{d}{9}

. Scale both bounds up to denominator

90

.

0.\overline{2} = \frac{2}{9} = \frac{20}{90}, \quad 0.\overline{7} = \frac{7}{9} = \frac{70}{90}

So the numerator

x

must satisfy

20 < x < 70

.

2Find which numerators terminate

#7 Identify Subproblems 8.NS.A.1

Factor the denominator:

90 = 2 \times 3^2 \times 5

. The

2

and

5

are harmless; the factor

3^2 = 9

must cancel completely, so

x

must be a multiple of

9

.

\frac{x}{90} = \frac{x}{2 \times 3^2 \times 5}

If any

3

is left in the denominator, the decimal repeats forever.

3List the multiples of 9 in range

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

The multiples of

9

strictly between

20

and

70

are

27

,

36

,

45

,

54

,

63

.

9 \times 3 = 27, \quad 9 \times 4 = 36, \quad 9 \times 5 = 45, \quad 9 \times 6 = 54, \quad 9 \times 7 = 63

The next ones,

18

and

72

, fall outside the range, so there are

5

.

Answer: 5

4 · Reviewdoes it hold up?

Check each: $\frac{27}{90} = \frac{3}{10} = 0.3$ , $\frac{36}{90} = \frac{2}{5} = 0.4$ , $\frac{45}{90} = \frac{1}{2} = 0.5$ , $\frac{54}{90} = \frac{3}{5} = 0.6$ , $\frac{63}{90} = \frac{7}{10} = 0.7$ . All terminate and all lie between $0.222\ldots$ and $0.777\ldots$ , while a neighbour such as $\frac{30}{90} = \frac{1}{3} = 0.\overline{3}$ does not terminate.

Another way: Count the complement: there are $49$ numerators from $21$ to $69$ ; every one except the $5$ multiples of $9$ leaves a factor $3$ in the denominator and repeats.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Turning the repeating bounds into fractions and deciding which fractions have terminating decimals.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that the numerator must be a multiple of $9$ and listing those multiples.

💡Takeaway. Put everything over one denominator, then just ask which numerators cancel the 3s.

Variant 6 medium

Among the fractions with denominator $90$ that lie between $0.\overline{5}$ and $0.\overline{8}$ , how many can be written as terminating decimals?

Your answer

Show solution

1 · Understandwhat's really being asked

Look at fractions $\frac{x}{90}$ that sit strictly between $0.\overline{5}$ and $0.\overline{8}$ . Count how many of them are terminating decimals.

Givens

The denominator is $90$ .
The fraction lies between $0.\overline{5}$ and $0.\overline{8}$ .

Unknowns

How many of these fractions are terminating decimals.

Constraints

A fraction in lowest terms terminates exactly when its denominator has no prime factor other than $2$ and $5$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#2 Make a Systematic List

The bounds are repeating decimals, so first rewrite them as fractions over $90$ ; that turns the question into a range of whole-number numerators. Then factor $90$ to see which numerators remove the troublesome prime $3$ , and list them.

3 · Execute3 carry out the plan

1Rewrite the bounds over 90

#15 Organize Information in More Ways 8.NS.A.1

A single repeating digit

d

after the point is

\frac{d}{9}

. Scale both bounds up to denominator

90

.

0.\overline{5} = \frac{5}{9} = \frac{50}{90}, \quad 0.\overline{8} = \frac{8}{9} = \frac{80}{90}

So the numerator

x

must satisfy

50 < x < 80

.

2Find which numerators terminate

#7 Identify Subproblems 8.NS.A.1

Factor the denominator:

90 = 2 \times 3^2 \times 5

. The

2

and

5

are harmless; the factor

3^2 = 9

must cancel completely, so

x

must be a multiple of

9

.

\frac{x}{90} = \frac{x}{2 \times 3^2 \times 5}

If any

3

is left in the denominator, the decimal repeats forever.

3List the multiples of 9 in range

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

The multiples of

9

strictly between

50

and

80

are

54

,

63

,

72

.

9 \times 6 = 54, \quad 9 \times 7 = 63, \quad 9 \times 8 = 72

The next ones,

45

and

81

, fall outside the range, so there are

3

.

Answer: 3

4 · Reviewdoes it hold up?

Check each: $\frac{54}{90} = \frac{3}{5} = 0.6$ , $\frac{63}{90} = \frac{7}{10} = 0.7$ , $\frac{72}{90} = \frac{4}{5} = 0.8$ . All terminate and all lie between $0.555\ldots$ and $0.888\ldots$ , while a neighbour such as $\frac{60}{90} = \frac{2}{3} = 0.\overline{6}$ does not terminate.

Another way: Count the complement: there are $29$ numerators from $51$ to $79$ ; every one except the $3$ multiples of $9$ leaves a factor $3$ in the denominator and repeats.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Turning the repeating bounds into fractions and deciding which fractions have terminating decimals.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that the numerator must be a multiple of $9$ and listing those multiples.

💡Takeaway. Put everything over one denominator, then just ask which numerators cancel the 3s.

Variant 7 medium

Among the fractions with denominator $90$ that lie between $0.\overline{1}$ and $0.\overline{4}$ , how many can be written as terminating decimals?

Your answer

Show solution

1 · Understandwhat's really being asked

Look at fractions $\frac{x}{90}$ that sit strictly between $0.\overline{1}$ and $0.\overline{4}$ . Count how many of them are terminating decimals.

Givens

The denominator is $90$ .
The fraction lies between $0.\overline{1}$ and $0.\overline{4}$ .

Unknowns

How many of these fractions are terminating decimals.

Constraints

A fraction in lowest terms terminates exactly when its denominator has no prime factor other than $2$ and $5$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#2 Make a Systematic List

The bounds are repeating decimals, so first rewrite them as fractions over $90$ ; that turns the question into a range of whole-number numerators. Then factor $90$ to see which numerators remove the troublesome prime $3$ , and list them.

3 · Execute3 carry out the plan

1Rewrite the bounds over 90

#15 Organize Information in More Ways 8.NS.A.1

A single repeating digit

d

after the point is

\frac{d}{9}

. Scale both bounds up to denominator

90

.

0.\overline{1} = \frac{1}{9} = \frac{10}{90}, \quad 0.\overline{4} = \frac{4}{9} = \frac{40}{90}

So the numerator

x

must satisfy

10 < x < 40

.

2Find which numerators terminate

#7 Identify Subproblems 8.NS.A.1

Factor the denominator:

90 = 2 \times 3^2 \times 5

. The

2

and

5

are harmless; the factor

3^2 = 9

must cancel completely, so

x

must be a multiple of

9

.

\frac{x}{90} = \frac{x}{2 \times 3^2 \times 5}

If any

3

is left in the denominator, the decimal repeats forever.

3List the multiples of 9 in range

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

The multiples of

9

strictly between

10

and

40

are

18

,

27

,

36

.

9 \times 2 = 18, \quad 9 \times 3 = 27, \quad 9 \times 4 = 36

The next ones,

9

and

45

, fall outside the range, so there are

3

.

Answer: 3

4 · Reviewdoes it hold up?

Check each: $\frac{18}{90} = \frac{1}{5} = 0.2$ , $\frac{27}{90} = \frac{3}{10} = 0.3$ , $\frac{36}{90} = \frac{2}{5} = 0.4$ . All terminate and all lie between $0.111\ldots$ and $0.444\ldots$ , while a neighbour such as $\frac{30}{90} = \frac{1}{3} = 0.\overline{3}$ does not terminate.

Another way: Count the complement: there are $29$ numerators from $11$ to $39$ ; every one except the $3$ multiples of $9$ leaves a factor $3$ in the denominator and repeats.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Turning the repeating bounds into fractions and deciding which fractions have terminating decimals.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that the numerator must be a multiple of $9$ and listing those multiples.

💡Takeaway. Put everything over one denominator, then just ask which numerators cancel the 3s.

Variant 8 medium

Among the fractions with denominator $90$ that lie between $0.\overline{3}$ and $0.\overline{6}$ , how many can be written as terminating decimals?

Your answer

Show solution

1 · Understandwhat's really being asked

Look at fractions $\frac{x}{90}$ that sit strictly between $0.\overline{3}$ and $0.\overline{6}$ . Count how many of them are terminating decimals.

Givens

The denominator is $90$ .
The fraction lies between $0.\overline{3}$ and $0.\overline{6}$ .

Unknowns

How many of these fractions are terminating decimals.

Constraints

A fraction in lowest terms terminates exactly when its denominator has no prime factor other than $2$ and $5$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#2 Make a Systematic List

The bounds are repeating decimals, so first rewrite them as fractions over $90$ ; that turns the question into a range of whole-number numerators. Then factor $90$ to see which numerators remove the troublesome prime $3$ , and list them.

3 · Execute3 carry out the plan

1Rewrite the bounds over 90

#15 Organize Information in More Ways 8.NS.A.1

A single repeating digit

d

after the point is

\frac{d}{9}

. Scale both bounds up to denominator

90

.

0.\overline{3} = \frac{3}{9} = \frac{30}{90}, \quad 0.\overline{6} = \frac{6}{9} = \frac{60}{90}

So the numerator

x

must satisfy

30 < x < 60

.

2Find which numerators terminate

#7 Identify Subproblems 8.NS.A.1

Factor the denominator:

90 = 2 \times 3^2 \times 5

. The

2

and

5

are harmless; the factor

3^2 = 9

must cancel completely, so

x

must be a multiple of

9

.

\frac{x}{90} = \frac{x}{2 \times 3^2 \times 5}

If any

3

is left in the denominator, the decimal repeats forever.

3List the multiples of 9 in range

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

The multiples of

9

strictly between

30

and

60

are

36

,

45

,

54

.

9 \times 4 = 36, \quad 9 \times 5 = 45, \quad 9 \times 6 = 54

The next ones,

27

and

63

, fall outside the range, so there are

3

.

Answer: 3

4 · Reviewdoes it hold up?

Check each: $\frac{36}{90} = \frac{2}{5} = 0.4$ , $\frac{45}{90} = \frac{1}{2} = 0.5$ , $\frac{54}{90} = \frac{3}{5} = 0.6$ . All terminate and all lie between $0.333\ldots$ and $0.666\ldots$ , while a neighbour such as $\frac{40}{90} = \frac{4}{9} = 0.\overline{4}$ does not terminate.

Another way: Count the complement: there are $29$ numerators from $31$ to $59$ ; every one except the $3$ multiples of $9$ leaves a factor $3$ in the denominator and repeats.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Turning the repeating bounds into fractions and deciding which fractions have terminating decimals.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that the numerator must be a multiple of $9$ and listing those multiples.

💡Takeaway. Put everything over one denominator, then just ask which numerators cancel the 3s.

Variant 9 hard

Among the fractions with denominator $108$ that lie between $0.\overline{3}$ and $0.\overline{8}$ , how many can be written as terminating decimals?

Your answer

Show solution

1 · Understandwhat's really being asked

Look at fractions $\frac{x}{108}$ that sit strictly between $0.\overline{3}$ and $0.\overline{8}$ . Count how many of them are terminating decimals.

Givens

The denominator is $108$ .
The fraction lies between $0.\overline{3}$ and $0.\overline{8}$ .

Unknowns

How many of these fractions are terminating decimals.

Constraints

A fraction in lowest terms terminates exactly when its denominator has no prime factor other than $2$ and $5$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#2 Make a Systematic List

The bounds are repeating decimals, so first rewrite them as fractions over $108$ ; that turns the question into a range of whole-number numerators. Then factor $108$ to see which numerators remove the troublesome prime $3$ , and list them.

3 · Execute3 carry out the plan

1Rewrite the bounds over 108

#15 Organize Information in More Ways 8.NS.A.1

A single repeating digit

d

after the point is

\frac{d}{9}

. Scale both bounds up to denominator

108

.

0.\overline{3} = \frac{3}{9} = \frac{36}{108}, \quad 0.\overline{8} = \frac{8}{9} = \frac{96}{108}

So the numerator

x

must satisfy

36 < x < 96

.

2Find which numerators terminate

#7 Identify Subproblems 8.NS.A.1

Factor the denominator:

108 = 2^2 \times 3^3

. The

2

is harmless; the factor

3^3 = 27

must cancel completely, so

x

must be a multiple of

27

.

\frac{x}{108} = \frac{x}{2^2 \times 3^3}

If any

3

is left in the denominator, the decimal repeats forever.

3List the multiples of 27 in range

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

The multiples of

27

strictly between

36

and

96

are

54

and

81

.

27 \times 2 = 54, \quad 27 \times 3 = 81

The next ones,

27

and

108

, fall outside the range, so there are

2

.

Answer: 2

4 · Reviewdoes it hold up?

Check each: $\frac{54}{108} = \frac{1}{2} = 0.5$ , $\frac{81}{108} = \frac{3}{4} = 0.75$ . All terminate and all lie between $0.333\ldots$ and $0.888\ldots$ , while a neighbour such as $\frac{72}{108} = \frac{2}{3} = 0.\overline{6}$ does not terminate.

Another way: Count the complement: there are $59$ numerators from $37$ to $95$ ; every one except the $2$ multiples of $27$ leaves a factor $3$ in the denominator and repeats.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Turning the repeating bounds into fractions and deciding which fractions have terminating decimals.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that the numerator must be a multiple of $27$ and listing those multiples.

💡Takeaway. Put everything over one denominator, then just ask which numerators cancel the 3s.

Variant 10 hard

Among the fractions with denominator $135$ that lie between $0.\overline{4}$ and $0.\overline{8}$ , how many can be written as terminating decimals?

Your answer

Show solution

1 · Understandwhat's really being asked

Look at fractions $\frac{x}{135}$ that sit strictly between $0.\overline{4}$ and $0.\overline{8}$ . Count how many of them are terminating decimals.

Givens

The denominator is $135$ .
The fraction lies between $0.\overline{4}$ and $0.\overline{8}$ .

Unknowns

How many of these fractions are terminating decimals.

Constraints

A fraction in lowest terms terminates exactly when its denominator has no prime factor other than $2$ and $5$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#2 Make a Systematic List

The bounds are repeating decimals, so first rewrite them as fractions over $135$ ; that turns the question into a range of whole-number numerators. Then factor $135$ to see which numerators remove the troublesome prime $3$ , and list them.

3 · Execute3 carry out the plan

1Rewrite the bounds over 135

#15 Organize Information in More Ways 8.NS.A.1

A single repeating digit

d

after the point is

\frac{d}{9}

. Scale both bounds up to denominator

135

.

0.\overline{4} = \frac{4}{9} = \frac{60}{135}, \quad 0.\overline{8} = \frac{8}{9} = \frac{120}{135}

So the numerator

x

must satisfy

60 < x < 120

.

2Find which numerators terminate

#7 Identify Subproblems 8.NS.A.1

Factor the denominator:

135 = 3^3 \times 5

. The

5

is harmless; the factor

3^3 = 27

must cancel completely, so

x

must be a multiple of

27

.

\frac{x}{135} = \frac{x}{3^3 \times 5}

If any

3

is left in the denominator, the decimal repeats forever.

3List the multiples of 27 in range

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

The multiples of

27

strictly between

60

and

120

are

81

and

108

.

27 \times 3 = 81, \quad 27 \times 4 = 108

The next ones,

54

and

135

, fall outside the range, so there are

2

.

Answer: 2

4 · Reviewdoes it hold up?

Check each: $\frac{81}{135} = \frac{3}{5} = 0.6$ , $\frac{108}{135} = \frac{4}{5} = 0.8$ . All terminate and all lie between $0.444\ldots$ and $0.888\ldots$ , while a neighbour such as $\frac{90}{135} = \frac{2}{3} = 0.\overline{6}$ does not terminate.

Another way: Count the complement: there are $59$ numerators from $61$ to $119$ ; every one except the $2$ multiples of $27$ leaves a factor $3$ in the denominator and repeats.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Turning the repeating bounds into fractions and deciding which fractions have terminating decimals.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that the numerator must be a multiple of $27$ and listing those multiples.

💡Takeaway. Put everything over one denominator, then just ask which numerators cancel the 3s.

Variant 11 hard

Among the fractions with denominator $135$ that lie between $0.\overline{2}$ and $0.\overline{8}$ , how many can be written as terminating decimals?

Your answer

Show solution

1 · Understandwhat's really being asked

Look at fractions $\frac{x}{135}$ that sit strictly between $0.\overline{2}$ and $0.\overline{8}$ . Count how many of them are terminating decimals.

Givens

The denominator is $135$ .
The fraction lies between $0.\overline{2}$ and $0.\overline{8}$ .

Unknowns

How many of these fractions are terminating decimals.

Constraints

A fraction in lowest terms terminates exactly when its denominator has no prime factor other than $2$ and $5$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#2 Make a Systematic List

The bounds are repeating decimals, so first rewrite them as fractions over $135$ ; that turns the question into a range of whole-number numerators. Then factor $135$ to see which numerators remove the troublesome prime $3$ , and list them.

3 · Execute3 carry out the plan

1Rewrite the bounds over 135

#15 Organize Information in More Ways 8.NS.A.1

A single repeating digit

d

after the point is

\frac{d}{9}

. Scale both bounds up to denominator

135

.

0.\overline{2} = \frac{2}{9} = \frac{30}{135}, \quad 0.\overline{8} = \frac{8}{9} = \frac{120}{135}

So the numerator

x

must satisfy

30 < x < 120

.

2Find which numerators terminate

#7 Identify Subproblems 8.NS.A.1

Factor the denominator:

135 = 3^3 \times 5

. The

5

is harmless; the factor

3^3 = 27

must cancel completely, so

x

must be a multiple of

27

.

\frac{x}{135} = \frac{x}{3^3 \times 5}

If any

3

is left in the denominator, the decimal repeats forever.

3List the multiples of 27 in range

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

The multiples of

27

strictly between

30

and

120

are

54

,

81

,

108

.

27 \times 2 = 54, \quad 27 \times 3 = 81, \quad 27 \times 4 = 108

The next ones,

27

and

135

, fall outside the range, so there are

3

.

Answer: 3

4 · Reviewdoes it hold up?

Check each: $\frac{54}{135} = \frac{2}{5} = 0.4$ , $\frac{81}{135} = \frac{3}{5} = 0.6$ , $\frac{108}{135} = \frac{4}{5} = 0.8$ . All terminate and all lie between $0.222\ldots$ and $0.888\ldots$ , while a neighbour such as $\frac{45}{135} = \frac{1}{3} = 0.\overline{3}$ does not terminate.

Another way: Count the complement: there are $89$ numerators from $31$ to $119$ ; every one except the $3$ multiples of $27$ leaves a factor $3$ in the denominator and repeats.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Turning the repeating bounds into fractions and deciding which fractions have terminating decimals.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that the numerator must be a multiple of $27$ and listing those multiples.

💡Takeaway. Put everything over one denominator, then just ask which numerators cancel the 3s.

Variant 12 hard

Among the fractions with denominator $144$ that lie between $0.\overline{3}$ and $0.\overline{5}$ , how many can be written as terminating decimals?

Your answer

Show solution

1 · Understandwhat's really being asked

Look at fractions $\frac{x}{144}$ that sit strictly between $0.\overline{3}$ and $0.\overline{5}$ . Count how many of them are terminating decimals.

Givens

The denominator is $144$ .
The fraction lies between $0.\overline{3}$ and $0.\overline{5}$ .

Unknowns

How many of these fractions are terminating decimals.

Constraints

A fraction in lowest terms terminates exactly when its denominator has no prime factor other than $2$ and $5$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#2 Make a Systematic List

The bounds are repeating decimals, so first rewrite them as fractions over $144$ ; that turns the question into a range of whole-number numerators. Then factor $144$ to see which numerators remove the troublesome prime $3$ , and list them.

3 · Execute3 carry out the plan

1Rewrite the bounds over 144

#15 Organize Information in More Ways 8.NS.A.1

A single repeating digit

d

after the point is

\frac{d}{9}

. Scale both bounds up to denominator

144

.

0.\overline{3} = \frac{3}{9} = \frac{48}{144}, \quad 0.\overline{5} = \frac{5}{9} = \frac{80}{144}

So the numerator

x

must satisfy

48 < x < 80

.

2Find which numerators terminate

#7 Identify Subproblems 8.NS.A.1

Factor the denominator:

144 = 2^4 \times 3^2

. The

2

is harmless; the factor

3^2 = 9

must cancel completely, so

x

must be a multiple of

9

.

\frac{x}{144} = \frac{x}{2^4 \times 3^2}

If any

3

is left in the denominator, the decimal repeats forever.

3List the multiples of 9 in range

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

The multiples of

9

strictly between

48

and

80

are

54

,

63

,

72

.

9 \times 6 = 54, \quad 9 \times 7 = 63, \quad 9 \times 8 = 72

The next ones,

45

and

81

, fall outside the range, so there are

3

.

Answer: 3

4 · Reviewdoes it hold up?

Check each: $\frac{54}{144} = \frac{3}{8} = 0.375$ , $\frac{63}{144} = \frac{7}{16} = 0.4375$ , $\frac{72}{144} = \frac{1}{2} = 0.5$ . All terminate and all lie between $0.333\ldots$ and $0.555\ldots$ , while a neighbour such as $\frac{64}{144} = \frac{4}{9} = 0.\overline{4}$ does not terminate.

Another way: Count the complement: there are $31$ numerators from $49$ to $79$ ; every one except the $3$ multiples of $9$ leaves a factor $3$ in the denominator and repeats.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Turning the repeating bounds into fractions and deciding which fractions have terminating decimals.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that the numerator must be a multiple of $9$ and listing those multiples.

💡Takeaway. Put everything over one denominator, then just ask which numerators cancel the 3s.

Count numerators that cancel the denominator's 3s · 12 practice problems

Generated variants — 12

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