Practice · Over 11 means the block is a multiple of 9 · Sensim Math

Variant 1 easy

When the repeating decimal $0.\overline{xy}$ (the two-digit block $xy$ repeats) is written as a fraction in lowest terms, it is $\frac{z}{11}$ . If $x$ is a one-digit odd number, how many natural numbers $z$ are possible?

Your answer

Show solution

1 · Understandwhat's really being asked

The repeating decimal $0.\overline{xy}$ equals $\frac{z}{11}$ in lowest terms, with $x$ a one-digit odd number. Count the possible natural numbers $z$ .

Givens

$0.\overline{xy} = \frac{z}{11}$ in lowest terms.
$x$ is a one-digit odd number: $1, 3, 5, 7, 9$ .

Unknowns

How many values $z$ can take.

Constraints

$y$ is a digit from $0$ to $9$ .
$\frac{z}{11}$ is in lowest terms, so $z$ is not a multiple of $11$ .

2 · Planchoose the strategy

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

Write both sides over the same denominator $99$ to get an equation in whole numbers. It says the two-digit number $xy$ is a multiple of $9$ , which a digit-sum list handles quickly.

3 · Execute4 carry out the plan

1Put both sides over 99

#13 Convert to Algebra 8.NS.A.1

A two-digit repeating block sits over

99

, and

\frac{z}{11} = \frac{9z}{99}

.

\frac{10x + y}{99} = \frac{9z}{99} \Rightarrow 10x + y = 9z

So the two-digit number

10x + y

must be a multiple of

9

.

2Use the digit-sum test

#13 Convert to Algebra 4.OA.B.4

A number is a multiple of

9

exactly when its digit sum is. With

x \le 9

and

y \le 9

, the digit sum

x + y

is

9

or

18

.

x + y = 9 \quad \text{or} \quad x + y = 18

Only two digit sums are possible.

3List the odd x

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

For

x = 1, 3, 5, 7, 9

with

x + y = 9

:

(1, 8)

,

(3, 6)

,

(5, 4)

,

(7, 2)

,

(9, 0)

. With

x + y = 18

the only pair is

(9, 9)

, but

0.\overline{99} = \frac{99}{99} = 1

and

\frac{11}{11}

is not in lowest terms, so it is out.

(x, y) = (1, 8), (3, 6), (5, 4), (7, 2), (9, 0)

Each pair gives a different block, so a different

z

.

4Read off z

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

Divide each block by

9

:

18, 36, 54, 72, 90

give

z = 2, 4, 6, 8, 10

. None is a multiple of

11

.

z = 2, 4, 6, 8, 10

There are

5

possible values.

Answer: 5

4 · Reviewdoes it hold up?

Spot-check: $0.\overline{36} = \frac{36}{99} = \frac{4}{11}$ and $0.\overline{90} = \frac{90}{99} = \frac{10}{11}$ , both in lowest terms with denominator $11$ .

Another way: List by $z$ instead: $9z$ for $z = 1$ to $10$ is $09, 18, 27, \ldots, 90$ ; the tens digit is odd exactly for $18, 36, 54, 72, 90$ , again $5$ values.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing the repeating decimal $0.\overline{xy}$ as the fraction $\frac{10x + y}{99}$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing multiples of $9$ by their digit sum and listing the cases.

💡Takeaway. Over 11 means over 99 -- so the repeating block has to be a multiple of 9.

Variant 2 easy

When the repeating decimal $0.\overline{xy}$ (the two-digit block $xy$ repeats) is written as a fraction in lowest terms, it is $\frac{z}{11}$ . If $x$ is a one-digit composite number, how many natural numbers $z$ are possible?

Your answer

Show solution

1 · Understandwhat's really being asked

The repeating decimal $0.\overline{xy}$ equals $\frac{z}{11}$ in lowest terms, with $x$ a one-digit composite number. Count the possible natural numbers $z$ .

Givens

$0.\overline{xy} = \frac{z}{11}$ in lowest terms.
$x$ is a one-digit composite number: $4, 6, 8, 9$ .

Unknowns

How many values $z$ can take.

Constraints

$y$ is a digit from $0$ to $9$ .
$\frac{z}{11}$ is in lowest terms, so $z$ is not a multiple of $11$ .

2 · Planchoose the strategy

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

Write both sides over the same denominator $99$ to get an equation in whole numbers. It says the two-digit number $xy$ is a multiple of $9$ , which a digit-sum list handles quickly.

3 · Execute4 carry out the plan

1Put both sides over 99

#13 Convert to Algebra 8.NS.A.1

A two-digit repeating block sits over

99

, and

\frac{z}{11} = \frac{9z}{99}

.

\frac{10x + y}{99} = \frac{9z}{99} \Rightarrow 10x + y = 9z

So the two-digit number

10x + y

must be a multiple of

9

.

2Use the digit-sum test

#13 Convert to Algebra 4.OA.B.4

A number is a multiple of

9

exactly when its digit sum is. With

x \le 9

and

y \le 9

, the digit sum

x + y

is

9

or

18

.

x + y = 9 \quad \text{or} \quad x + y = 18

Only two digit sums are possible.

3List the composite x

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

For

x = 4, 6, 8, 9

with

x + y = 9

:

(4, 5)

,

(6, 3)

,

(8, 1)

,

(9, 0)

. With

x + y = 18

the only pair is

(9, 9)

, but

0.\overline{99} = \frac{99}{99} = 1

and

\frac{11}{11}

is not in lowest terms, so it is out.

(x, y) = (4, 5), (6, 3), (8, 1), (9, 0)

Each pair gives a different block, so a different

z

.

4Read off z

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

Divide each block by

9

:

45, 63, 81, 90

give

z = 5, 7, 9, 10

. None is a multiple of

11

.

z = 5, 7, 9, 10

There are

4

possible values.

Answer: 4

4 · Reviewdoes it hold up?

Spot-check: $0.\overline{63} = \frac{63}{99} = \frac{7}{11}$ and $0.\overline{90} = \frac{90}{99} = \frac{10}{11}$ , both in lowest terms with denominator $11$ .

Another way: List by $z$ instead: $9z$ for $z = 1$ to $10$ is $09, 18, 27, \ldots, 90$ ; the tens digit is composite exactly for $45, 63, 81, 90$ , again $4$ values.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing the repeating decimal $0.\overline{xy}$ as the fraction $\frac{10x + y}{99}$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing multiples of $9$ by their digit sum and listing the cases.

💡Takeaway. Over 11 means over 99 -- so the repeating block has to be a multiple of 9.

Variant 3 hard

When the repeating decimal $0.\overline{xy}$ (the two-digit block $xy$ repeats) is written as a fraction in lowest terms, it is $\frac{z}{11}$ . If $x$ is a one-digit number greater than $7$ , how many natural numbers $z$ are possible?

Your answer

Show solution

1 · Understandwhat's really being asked

The repeating decimal $0.\overline{xy}$ equals $\frac{z}{11}$ in lowest terms, with $x$ a one-digit number greater than $7$ . Count the possible natural numbers $z$ .

Givens

$0.\overline{xy} = \frac{z}{11}$ in lowest terms.
$x$ is a one-digit number greater than $7$ : $8, 9$ .

Unknowns

How many values $z$ can take.

Constraints

$y$ is a digit from $0$ to $9$ .
$\frac{z}{11}$ is in lowest terms, so $z$ is not a multiple of $11$ .

2 · Planchoose the strategy

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

Write both sides over the same denominator $99$ to get an equation in whole numbers. It says the two-digit number $xy$ is a multiple of $9$ , which a digit-sum list handles quickly.

3 · Execute4 carry out the plan

1Put both sides over 99

#13 Convert to Algebra 8.NS.A.1

A two-digit repeating block sits over

99

, and

\frac{z}{11} = \frac{9z}{99}

.

\frac{10x + y}{99} = \frac{9z}{99} \Rightarrow 10x + y = 9z

So the two-digit number

10x + y

must be a multiple of

9

.

2Use the digit-sum test

#13 Convert to Algebra 4.OA.B.4

A number is a multiple of

9

exactly when its digit sum is. With

x \le 9

and

y \le 9

, the digit sum

x + y

is

9

or

18

.

x + y = 9 \quad \text{or} \quad x + y = 18

Only two digit sums are possible.

3List the allowed x

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

For

x = 8, 9

with

x + y = 9

:

(8, 1)

,

(9, 0)

. With

x + y = 18

the only pair is

(9, 9)

, but

0.\overline{99} = \frac{99}{99} = 1

and

\frac{11}{11}

is not in lowest terms, so it is out.

(x, y) = (8, 1), (9, 0)

Each pair gives a different block, so a different

z

.

4Read off z

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

Divide each block by

9

:

81, 90

give

z = 9, 10

. None is a multiple of

11

.

z = 9, 10

There are

2

possible values.

Answer: 2

4 · Reviewdoes it hold up?

Spot-check: $0.\overline{81} = \frac{81}{99} = \frac{9}{11}$ and $0.\overline{90} = \frac{90}{99} = \frac{10}{11}$ , both in lowest terms with denominator $11$ .

Another way: List by $z$ instead: $9z$ for $z = 1$ to $10$ is $09, 18, 27, \ldots, 90$ ; the tens digit is greater than $7$ exactly for $81, 90$ , again $2$ values.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing the repeating decimal $0.\overline{xy}$ as the fraction $\frac{10x + y}{99}$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing multiples of $9$ by their digit sum and listing the cases.

💡Takeaway. Over 11 means over 99 -- so the repeating block has to be a multiple of 9.

Variant 4 medium

When the repeating decimal $0.\overline{xy}$ (the two-digit block $xy$ repeats) is written as a fraction in lowest terms, it is $\frac{z}{11}$ . If $x$ is a one-digit number greater than $3$ , how many natural numbers $z$ are possible?

Your answer

Show solution

1 · Understandwhat's really being asked

The repeating decimal $0.\overline{xy}$ equals $\frac{z}{11}$ in lowest terms, with $x$ a one-digit number greater than $3$ . Count the possible natural numbers $z$ .

Givens

$0.\overline{xy} = \frac{z}{11}$ in lowest terms.
$x$ is a one-digit number greater than $3$ : $4, 5, 6, 7, 8, 9$ .

Unknowns

How many values $z$ can take.

Constraints

$y$ is a digit from $0$ to $9$ .
$\frac{z}{11}$ is in lowest terms, so $z$ is not a multiple of $11$ .

2 · Planchoose the strategy

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

Write both sides over the same denominator $99$ to get an equation in whole numbers. It says the two-digit number $xy$ is a multiple of $9$ , which a digit-sum list handles quickly.

3 · Execute4 carry out the plan

1Put both sides over 99

#13 Convert to Algebra 8.NS.A.1

A two-digit repeating block sits over

99

, and

\frac{z}{11} = \frac{9z}{99}

.

\frac{10x + y}{99} = \frac{9z}{99} \Rightarrow 10x + y = 9z

So the two-digit number

10x + y

must be a multiple of

9

.

2Use the digit-sum test

#13 Convert to Algebra 4.OA.B.4

A number is a multiple of

9

exactly when its digit sum is. With

x \le 9

and

y \le 9

, the digit sum

x + y

is

9

or

18

.

x + y = 9 \quad \text{or} \quad x + y = 18

Only two digit sums are possible.

3List the allowed x

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

For

x = 4, 5, 6, 7, 8, 9

with

x + y = 9

:

(4, 5)

,

(5, 4)

,

(6, 3)

,

(7, 2)

,

(8, 1)

,

(9, 0)

. With

x + y = 18

the only pair is

(9, 9)

, but

0.\overline{99} = \frac{99}{99} = 1

and

\frac{11}{11}

is not in lowest terms, so it is out.

(x, y) = (4, 5), (5, 4), (6, 3), (7, 2), (8, 1), (9, 0)

Each pair gives a different block, so a different

z

.

4Read off z

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

Divide each block by

9

:

45, 54, 63, 72, 81, 90

give

z = 5, 6, 7, 8, 9, 10

. None is a multiple of

11

.

z = 5, 6, 7, 8, 9, 10

There are

6

possible values.

Answer: 6

4 · Reviewdoes it hold up?

Spot-check: $0.\overline{54} = \frac{54}{99} = \frac{6}{11}$ and $0.\overline{90} = \frac{90}{99} = \frac{10}{11}$ , both in lowest terms with denominator $11$ .

Another way: List by $z$ instead: $9z$ for $z = 1$ to $10$ is $09, 18, 27, \ldots, 90$ ; the tens digit is greater than $3$ exactly for $45, 54, 63, 72, 81, 90$ , again $6$ values.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing the repeating decimal $0.\overline{xy}$ as the fraction $\frac{10x + y}{99}$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing multiples of $9$ by their digit sum and listing the cases.

💡Takeaway. Over 11 means over 99 -- so the repeating block has to be a multiple of 9.

Variant 5 easy

When the repeating decimal $0.\overline{xy}$ (the two-digit block $xy$ repeats) is written as a fraction in lowest terms, it is $\frac{z}{11}$ . If $x$ is a one-digit number greater than $2$ , how many natural numbers $z$ are possible?

Your answer

Show solution

1 · Understandwhat's really being asked

The repeating decimal $0.\overline{xy}$ equals $\frac{z}{11}$ in lowest terms, with $x$ a one-digit number greater than $2$ . Count the possible natural numbers $z$ .

Givens

$0.\overline{xy} = \frac{z}{11}$ in lowest terms.
$x$ is a one-digit number greater than $2$ : $3, 4, 5, 6, 7, 8, 9$ .

Unknowns

How many values $z$ can take.

Constraints

$y$ is a digit from $0$ to $9$ .
$\frac{z}{11}$ is in lowest terms, so $z$ is not a multiple of $11$ .

2 · Planchoose the strategy

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

Write both sides over the same denominator $99$ to get an equation in whole numbers. It says the two-digit number $xy$ is a multiple of $9$ , which a digit-sum list handles quickly.

3 · Execute4 carry out the plan

1Put both sides over 99

#13 Convert to Algebra 8.NS.A.1

A two-digit repeating block sits over

99

, and

\frac{z}{11} = \frac{9z}{99}

.

\frac{10x + y}{99} = \frac{9z}{99} \Rightarrow 10x + y = 9z

So the two-digit number

10x + y

must be a multiple of

9

.

2Use the digit-sum test

#13 Convert to Algebra 4.OA.B.4

A number is a multiple of

9

exactly when its digit sum is. With

x \le 9

and

y \le 9

, the digit sum

x + y

is

9

or

18

.

x + y = 9 \quad \text{or} \quad x + y = 18

Only two digit sums are possible.

3List the allowed x

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

For

x = 3, 4, 5, 6, 7, 8, 9

with

x + y = 9

:

(3, 6)

,

(4, 5)

,

(5, 4)

,

(6, 3)

,

(7, 2)

,

(8, 1)

,

(9, 0)

. With

x + y = 18

the only pair is

(9, 9)

, but

0.\overline{99} = \frac{99}{99} = 1

and

\frac{11}{11}

is not in lowest terms, so it is out.

(x, y) = (3, 6), (4, 5), (5, 4), (6, 3), (7, 2), (8, 1), (9, 0)

Each pair gives a different block, so a different

z

.

4Read off z

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

Divide each block by

9

:

36, 45, 54, 63, 72, 81, 90

give

z = 4, 5, 6, 7, 8, 9, 10

. None is a multiple of

11

.

z = 4, 5, 6, 7, 8, 9, 10

There are

7

possible values.

Answer: 7

4 · Reviewdoes it hold up?

Spot-check: $0.\overline{45} = \frac{45}{99} = \frac{5}{11}$ and $0.\overline{90} = \frac{90}{99} = \frac{10}{11}$ , both in lowest terms with denominator $11$ .

Another way: List by $z$ instead: $9z$ for $z = 1$ to $10$ is $09, 18, 27, \ldots, 90$ ; the tens digit is greater than $2$ exactly for $36, 45, 54, 63, 72, 81, 90$ , again $7$ values.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing the repeating decimal $0.\overline{xy}$ as the fraction $\frac{10x + y}{99}$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing multiples of $9$ by their digit sum and listing the cases.

💡Takeaway. Over 11 means over 99 -- so the repeating block has to be a multiple of 9.

Variant 6 hard

When the repeating decimal $0.\overline{xy}$ (the two-digit block $xy$ repeats) is written as a fraction in lowest terms, it is $\frac{z}{11}$ . If $x$ is a one-digit odd number greater than $3$ , how many natural numbers $z$ are possible?

Your answer

Show solution

1 · Understandwhat's really being asked

The repeating decimal $0.\overline{xy}$ equals $\frac{z}{11}$ in lowest terms, with $x$ a one-digit odd number greater than $3$ . Count the possible natural numbers $z$ .

Givens

$0.\overline{xy} = \frac{z}{11}$ in lowest terms.
$x$ is a one-digit odd number greater than $3$ : $5, 7, 9$ .

Unknowns

How many values $z$ can take.

Constraints

$y$ is a digit from $0$ to $9$ .
$\frac{z}{11}$ is in lowest terms, so $z$ is not a multiple of $11$ .

2 · Planchoose the strategy

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

Write both sides over the same denominator $99$ to get an equation in whole numbers. It says the two-digit number $xy$ is a multiple of $9$ , which a digit-sum list handles quickly.

3 · Execute4 carry out the plan

1Put both sides over 99

#13 Convert to Algebra 8.NS.A.1

A two-digit repeating block sits over

99

, and

\frac{z}{11} = \frac{9z}{99}

.

\frac{10x + y}{99} = \frac{9z}{99} \Rightarrow 10x + y = 9z

So the two-digit number

10x + y

must be a multiple of

9

.

2Use the digit-sum test

#13 Convert to Algebra 4.OA.B.4

A number is a multiple of

9

exactly when its digit sum is. With

x \le 9

and

y \le 9

, the digit sum

x + y

is

9

or

18

.

x + y = 9 \quad \text{or} \quad x + y = 18

Only two digit sums are possible.

3List the allowed x

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

For

x = 5, 7, 9

with

x + y = 9

:

(5, 4)

,

(7, 2)

,

(9, 0)

. With

x + y = 18

the only pair is

(9, 9)

, but

0.\overline{99} = \frac{99}{99} = 1

and

\frac{11}{11}

is not in lowest terms, so it is out.

(x, y) = (5, 4), (7, 2), (9, 0)

Each pair gives a different block, so a different

z

.

4Read off z

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

Divide each block by

9

:

54, 72, 90

give

z = 6, 8, 10

. None is a multiple of

11

.

z = 6, 8, 10

There are

3

possible values.

Answer: 3

4 · Reviewdoes it hold up?

Spot-check: $0.\overline{72} = \frac{72}{99} = \frac{8}{11}$ and $0.\overline{90} = \frac{90}{99} = \frac{10}{11}$ , both in lowest terms with denominator $11$ .

Another way: List by $z$ instead: $9z$ for $z = 1$ to $10$ is $09, 18, 27, \ldots, 90$ ; the tens digit is odd and greater than $3$ exactly for $54, 72, 90$ , again $3$ values.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing the repeating decimal $0.\overline{xy}$ as the fraction $\frac{10x + y}{99}$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing multiples of $9$ by their digit sum and listing the cases.

💡Takeaway. Over 11 means over 99 -- so the repeating block has to be a multiple of 9.

Variant 7 hard

When the repeating decimal $0.\overline{xy}$ (the two-digit block $xy$ repeats) is written as a fraction in lowest terms, it is $\frac{z}{11}$ . If $x$ is a one-digit odd number greater than $5$ , how many natural numbers $z$ are possible?

Your answer

Show solution

1 · Understandwhat's really being asked

The repeating decimal $0.\overline{xy}$ equals $\frac{z}{11}$ in lowest terms, with $x$ a one-digit odd number greater than $5$ . Count the possible natural numbers $z$ .

Givens

$0.\overline{xy} = \frac{z}{11}$ in lowest terms.
$x$ is a one-digit odd number greater than $5$ : $7, 9$ .

Unknowns

How many values $z$ can take.

Constraints

$y$ is a digit from $0$ to $9$ .
$\frac{z}{11}$ is in lowest terms, so $z$ is not a multiple of $11$ .

2 · Planchoose the strategy

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

Write both sides over the same denominator $99$ to get an equation in whole numbers. It says the two-digit number $xy$ is a multiple of $9$ , which a digit-sum list handles quickly.

3 · Execute4 carry out the plan

1Put both sides over 99

#13 Convert to Algebra 8.NS.A.1

A two-digit repeating block sits over

99

, and

\frac{z}{11} = \frac{9z}{99}

.

\frac{10x + y}{99} = \frac{9z}{99} \Rightarrow 10x + y = 9z

So the two-digit number

10x + y

must be a multiple of

9

.

2Use the digit-sum test

#13 Convert to Algebra 4.OA.B.4

A number is a multiple of

9

exactly when its digit sum is. With

x \le 9

and

y \le 9

, the digit sum

x + y

is

9

or

18

.

x + y = 9 \quad \text{or} \quad x + y = 18

Only two digit sums are possible.

3List the allowed x

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

For

x = 7, 9

with

x + y = 9

:

(7, 2)

,

(9, 0)

. With

x + y = 18

the only pair is

(9, 9)

, but

0.\overline{99} = \frac{99}{99} = 1

and

\frac{11}{11}

is not in lowest terms, so it is out.

(x, y) = (7, 2), (9, 0)

Each pair gives a different block, so a different

z

.

4Read off z

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

Divide each block by

9

:

72, 90

give

z = 8, 10

. None is a multiple of

11

.

z = 8, 10

There are

2

possible values.

Answer: 2

4 · Reviewdoes it hold up?

Spot-check: $0.\overline{72} = \frac{72}{99} = \frac{8}{11}$ and $0.\overline{90} = \frac{90}{99} = \frac{10}{11}$ , both in lowest terms with denominator $11$ .

Another way: List by $z$ instead: $9z$ for $z = 1$ to $10$ is $09, 18, 27, \ldots, 90$ ; the tens digit is odd and greater than $5$ exactly for $72, 90$ , again $2$ values.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing the repeating decimal $0.\overline{xy}$ as the fraction $\frac{10x + y}{99}$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing multiples of $9$ by their digit sum and listing the cases.

💡Takeaway. Over 11 means over 99 -- so the repeating block has to be a multiple of 9.

Variant 8 medium

When the repeating decimal $0.\overline{xy}$ (the two-digit block $xy$ repeats) is written as a fraction in lowest terms, it is $\frac{z}{11}$ . If $x$ is a one-digit number greater than $5$ , how many natural numbers $z$ are possible?

Your answer

Show solution

1 · Understandwhat's really being asked

The repeating decimal $0.\overline{xy}$ equals $\frac{z}{11}$ in lowest terms, with $x$ a one-digit number greater than $5$ . Count the possible natural numbers $z$ .

Givens

$0.\overline{xy} = \frac{z}{11}$ in lowest terms.
$x$ is a one-digit number greater than $5$ : $6, 7, 8, 9$ .

Unknowns

How many values $z$ can take.

Constraints

$y$ is a digit from $0$ to $9$ .
$\frac{z}{11}$ is in lowest terms, so $z$ is not a multiple of $11$ .

2 · Planchoose the strategy

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

Write both sides over the same denominator $99$ to get an equation in whole numbers. It says the two-digit number $xy$ is a multiple of $9$ , which a digit-sum list handles quickly.

3 · Execute4 carry out the plan

1Put both sides over 99

#13 Convert to Algebra 8.NS.A.1

A two-digit repeating block sits over

99

, and

\frac{z}{11} = \frac{9z}{99}

.

\frac{10x + y}{99} = \frac{9z}{99} \Rightarrow 10x + y = 9z

So the two-digit number

10x + y

must be a multiple of

9

.

2Use the digit-sum test

#13 Convert to Algebra 4.OA.B.4

A number is a multiple of

9

exactly when its digit sum is. With

x \le 9

and

y \le 9

, the digit sum

x + y

is

9

or

18

.

x + y = 9 \quad \text{or} \quad x + y = 18

Only two digit sums are possible.

3List the allowed x

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

For

x = 6, 7, 8, 9

with

x + y = 9

:

(6, 3)

,

(7, 2)

,

(8, 1)

,

(9, 0)

. With

x + y = 18

the only pair is

(9, 9)

, but

0.\overline{99} = \frac{99}{99} = 1

and

\frac{11}{11}

is not in lowest terms, so it is out.

(x, y) = (6, 3), (7, 2), (8, 1), (9, 0)

Each pair gives a different block, so a different

z

.

4Read off z

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

Divide each block by

9

:

63, 72, 81, 90

give

z = 7, 8, 9, 10

. None is a multiple of

11

.

z = 7, 8, 9, 10

There are

4

possible values.

Answer: 4

4 · Reviewdoes it hold up?

Spot-check: $0.\overline{72} = \frac{72}{99} = \frac{8}{11}$ and $0.\overline{90} = \frac{90}{99} = \frac{10}{11}$ , both in lowest terms with denominator $11$ .

Another way: List by $z$ instead: $9z$ for $z = 1$ to $10$ is $09, 18, 27, \ldots, 90$ ; the tens digit is greater than $5$ exactly for $63, 72, 81, 90$ , again $4$ values.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing the repeating decimal $0.\overline{xy}$ as the fraction $\frac{10x + y}{99}$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing multiples of $9$ by their digit sum and listing the cases.

💡Takeaway. Over 11 means over 99 -- so the repeating block has to be a multiple of 9.

Variant 9 easy

When the repeating decimal $0.\overline{xy}$ (the two-digit block $xy$ repeats) is written as a fraction in lowest terms, it is $\frac{z}{11}$ . If $x$ is a one-digit natural number that is a multiple of $3$ , how many natural numbers $z$ are possible?

Your answer

Show solution

1 · Understandwhat's really being asked

The repeating decimal $0.\overline{xy}$ equals $\frac{z}{11}$ in lowest terms, with $x$ a one-digit natural number that is a multiple of $3$ . Count the possible natural numbers $z$ .

Givens

$0.\overline{xy} = \frac{z}{11}$ in lowest terms.
$x$ is a one-digit natural number that is a multiple of $3$ : $3, 6, 9$ .

Unknowns

How many values $z$ can take.

Constraints

$y$ is a digit from $0$ to $9$ .
$\frac{z}{11}$ is in lowest terms, so $z$ is not a multiple of $11$ .

2 · Planchoose the strategy

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

Write both sides over the same denominator $99$ to get an equation in whole numbers. It says the two-digit number $xy$ is a multiple of $9$ , which a digit-sum list handles quickly.

3 · Execute4 carry out the plan

1Put both sides over 99

#13 Convert to Algebra 8.NS.A.1

A two-digit repeating block sits over

99

, and

\frac{z}{11} = \frac{9z}{99}

.

\frac{10x + y}{99} = \frac{9z}{99} \Rightarrow 10x + y = 9z

So the two-digit number

10x + y

must be a multiple of

9

.

2Use the digit-sum test

#13 Convert to Algebra 4.OA.B.4

A number is a multiple of

9

exactly when its digit sum is. With

x \le 9

and

y \le 9

, the digit sum

x + y

is

9

or

18

.

x + y = 9 \quad \text{or} \quad x + y = 18

Only two digit sums are possible.

3List the allowed x

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

For

x = 3, 6, 9

with

x + y = 9

:

(3, 6)

,

(6, 3)

,

(9, 0)

. With

x + y = 18

the only pair is

(9, 9)

, but

0.\overline{99} = \frac{99}{99} = 1

and

\frac{11}{11}

is not in lowest terms, so it is out.

(x, y) = (3, 6), (6, 3), (9, 0)

Each pair gives a different block, so a different

z

.

4Read off z

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

Divide each block by

9

:

36, 63, 90

give

z = 4, 7, 10

. None is a multiple of

11

.

z = 4, 7, 10

There are

3

possible values.

Answer: 3

4 · Reviewdoes it hold up?

Spot-check: $0.\overline{63} = \frac{63}{99} = \frac{7}{11}$ and $0.\overline{90} = \frac{90}{99} = \frac{10}{11}$ , both in lowest terms with denominator $11$ .

Another way: List by $z$ instead: $9z$ for $z = 1$ to $10$ is $09, 18, 27, \ldots, 90$ ; the tens digit is a nonzero multiple of $3$ exactly for $36, 63, 90$ , again $3$ values.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing the repeating decimal $0.\overline{xy}$ as the fraction $\frac{10x + y}{99}$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing multiples of $9$ by their digit sum and listing the cases.

💡Takeaway. Over 11 means over 99 -- so the repeating block has to be a multiple of 9.

Variant 10 hard

When the repeating decimal $0.\overline{xy}$ (the two-digit block $xy$ repeats) is written as a fraction in lowest terms, it is $\frac{z}{11}$ . If $x$ is a one-digit odd number greater than $1$ , how many natural numbers $z$ are possible?

Your answer

Show solution

1 · Understandwhat's really being asked

The repeating decimal $0.\overline{xy}$ equals $\frac{z}{11}$ in lowest terms, with $x$ a one-digit odd number greater than $1$ . Count the possible natural numbers $z$ .

Givens

$0.\overline{xy} = \frac{z}{11}$ in lowest terms.
$x$ is a one-digit odd number greater than $1$ : $3, 5, 7, 9$ .

Unknowns

How many values $z$ can take.

Constraints

$y$ is a digit from $0$ to $9$ .
$\frac{z}{11}$ is in lowest terms, so $z$ is not a multiple of $11$ .

2 · Planchoose the strategy

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

Write both sides over the same denominator $99$ to get an equation in whole numbers. It says the two-digit number $xy$ is a multiple of $9$ , which a digit-sum list handles quickly.

3 · Execute4 carry out the plan

1Put both sides over 99

#13 Convert to Algebra 8.NS.A.1

A two-digit repeating block sits over

99

, and

\frac{z}{11} = \frac{9z}{99}

.

\frac{10x + y}{99} = \frac{9z}{99} \Rightarrow 10x + y = 9z

So the two-digit number

10x + y

must be a multiple of

9

.

2Use the digit-sum test

#13 Convert to Algebra 4.OA.B.4

A number is a multiple of

9

exactly when its digit sum is. With

x \le 9

and

y \le 9

, the digit sum

x + y

is

9

or

18

.

x + y = 9 \quad \text{or} \quad x + y = 18

Only two digit sums are possible.

3List the allowed x

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

For

x = 3, 5, 7, 9

with

x + y = 9

:

(3, 6)

,

(5, 4)

,

(7, 2)

,

(9, 0)

. With

x + y = 18

the only pair is

(9, 9)

, but

0.\overline{99} = \frac{99}{99} = 1

and

\frac{11}{11}

is not in lowest terms, so it is out.

(x, y) = (3, 6), (5, 4), (7, 2), (9, 0)

Each pair gives a different block, so a different

z

.

4Read off z

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

Divide each block by

9

:

36, 54, 72, 90

give

z = 4, 6, 8, 10

. None is a multiple of

11

.

z = 4, 6, 8, 10

There are

4

possible values.

Answer: 4

4 · Reviewdoes it hold up?

Spot-check: $0.\overline{54} = \frac{54}{99} = \frac{6}{11}$ and $0.\overline{90} = \frac{90}{99} = \frac{10}{11}$ , both in lowest terms with denominator $11$ .

Another way: List by $z$ instead: $9z$ for $z = 1$ to $10$ is $09, 18, 27, \ldots, 90$ ; the tens digit is odd and greater than $1$ exactly for $36, 54, 72, 90$ , again $4$ values.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing the repeating decimal $0.\overline{xy}$ as the fraction $\frac{10x + y}{99}$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing multiples of $9$ by their digit sum and listing the cases.

💡Takeaway. Over 11 means over 99 -- so the repeating block has to be a multiple of 9.

Variant 11 medium

When the repeating decimal $0.\overline{xy}$ (the two-digit block $xy$ repeats) is written as a fraction in lowest terms, it is $\frac{z}{11}$ . If $x$ is a one-digit number greater than $6$ , how many natural numbers $z$ are possible?

Your answer

Show solution

1 · Understandwhat's really being asked

The repeating decimal $0.\overline{xy}$ equals $\frac{z}{11}$ in lowest terms, with $x$ a one-digit number greater than $6$ . Count the possible natural numbers $z$ .

Givens

$0.\overline{xy} = \frac{z}{11}$ in lowest terms.
$x$ is a one-digit number greater than $6$ : $7, 8, 9$ .

Unknowns

How many values $z$ can take.

Constraints

$y$ is a digit from $0$ to $9$ .
$\frac{z}{11}$ is in lowest terms, so $z$ is not a multiple of $11$ .

2 · Planchoose the strategy

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

Write both sides over the same denominator $99$ to get an equation in whole numbers. It says the two-digit number $xy$ is a multiple of $9$ , which a digit-sum list handles quickly.

3 · Execute4 carry out the plan

1Put both sides over 99

#13 Convert to Algebra 8.NS.A.1

A two-digit repeating block sits over

99

, and

\frac{z}{11} = \frac{9z}{99}

.

\frac{10x + y}{99} = \frac{9z}{99} \Rightarrow 10x + y = 9z

So the two-digit number

10x + y

must be a multiple of

9

.

2Use the digit-sum test

#13 Convert to Algebra 4.OA.B.4

A number is a multiple of

9

exactly when its digit sum is. With

x \le 9

and

y \le 9

, the digit sum

x + y

is

9

or

18

.

x + y = 9 \quad \text{or} \quad x + y = 18

Only two digit sums are possible.

3List the allowed x

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

For

x = 7, 8, 9

with

x + y = 9

:

(7, 2)

,

(8, 1)

,

(9, 0)

. With

x + y = 18

the only pair is

(9, 9)

, but

0.\overline{99} = \frac{99}{99} = 1

and

\frac{11}{11}

is not in lowest terms, so it is out.

(x, y) = (7, 2), (8, 1), (9, 0)

Each pair gives a different block, so a different

z

.

4Read off z

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

Divide each block by

9

:

72, 81, 90

give

z = 8, 9, 10

. None is a multiple of

11

.

z = 8, 9, 10

There are

3

possible values.

Answer: 3

4 · Reviewdoes it hold up?

Spot-check: $0.\overline{81} = \frac{81}{99} = \frac{9}{11}$ and $0.\overline{90} = \frac{90}{99} = \frac{10}{11}$ , both in lowest terms with denominator $11$ .

Another way: List by $z$ instead: $9z$ for $z = 1$ to $10$ is $09, 18, 27, \ldots, 90$ ; the tens digit is greater than $6$ exactly for $72, 81, 90$ , again $3$ values.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing the repeating decimal $0.\overline{xy}$ as the fraction $\frac{10x + y}{99}$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing multiples of $9$ by their digit sum and listing the cases.

💡Takeaway. Over 11 means over 99 -- so the repeating block has to be a multiple of 9.

Variant 12 medium

When the repeating decimal $0.\overline{xy}$ (the two-digit block $xy$ repeats) is written as a fraction in lowest terms, it is $\frac{z}{11}$ . If $x$ is a one-digit number greater than $4$ , how many natural numbers $z$ are possible?

Your answer

Show solution

1 · Understandwhat's really being asked

The repeating decimal $0.\overline{xy}$ equals $\frac{z}{11}$ in lowest terms, with $x$ a one-digit number greater than $4$ . Count the possible natural numbers $z$ .

Givens

$0.\overline{xy} = \frac{z}{11}$ in lowest terms.
$x$ is a one-digit number greater than $4$ : $5, 6, 7, 8, 9$ .

Unknowns

How many values $z$ can take.

Constraints

$y$ is a digit from $0$ to $9$ .
$\frac{z}{11}$ is in lowest terms, so $z$ is not a multiple of $11$ .

2 · Planchoose the strategy

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

Write both sides over the same denominator $99$ to get an equation in whole numbers. It says the two-digit number $xy$ is a multiple of $9$ , which a digit-sum list handles quickly.

3 · Execute4 carry out the plan

1Put both sides over 99

#13 Convert to Algebra 8.NS.A.1

A two-digit repeating block sits over

99

, and

\frac{z}{11} = \frac{9z}{99}

.

\frac{10x + y}{99} = \frac{9z}{99} \Rightarrow 10x + y = 9z

So the two-digit number

10x + y

must be a multiple of

9

.

2Use the digit-sum test

#13 Convert to Algebra 4.OA.B.4

A number is a multiple of

9

exactly when its digit sum is. With

x \le 9

and

y \le 9

, the digit sum

x + y

is

9

or

18

.

x + y = 9 \quad \text{or} \quad x + y = 18

Only two digit sums are possible.

3List the allowed x

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

For

x = 5, 6, 7, 8, 9

with

x + y = 9

:

(5, 4)

,

(6, 3)

,

(7, 2)

,

(8, 1)

,

(9, 0)

. With

x + y = 18

the only pair is

(9, 9)

, but

0.\overline{99} = \frac{99}{99} = 1

and

\frac{11}{11}

is not in lowest terms, so it is out.

(x, y) = (5, 4), (6, 3), (7, 2), (8, 1), (9, 0)

Each pair gives a different block, so a different

z

.

4Read off z

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

Divide each block by

9

:

54, 63, 72, 81, 90

give

z = 6, 7, 8, 9, 10

. None is a multiple of

11

.

z = 6, 7, 8, 9, 10

There are

5

possible values.

Answer: 5

4 · Reviewdoes it hold up?

Spot-check: $0.\overline{63} = \frac{63}{99} = \frac{7}{11}$ and $0.\overline{90} = \frac{90}{99} = \frac{10}{11}$ , both in lowest terms with denominator $11$ .

Another way: List by $z$ instead: $9z$ for $z = 1$ to $10$ is $09, 18, 27, \ldots, 90$ ; the tens digit is greater than $4$ exactly for $54, 63, 72, 81, 90$ , again $5$ values.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing the repeating decimal $0.\overline{xy}$ as the fraction $\frac{10x + y}{99}$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing multiples of $9$ by their digit sum and listing the cases.

💡Takeaway. Over 11 means over 99 -- so the repeating block has to be a multiple of 9.

Over 11 means the block is a multiple of 9 · 12 practice problems

Generated variants — 12

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Where next?