Practice · Reversed repeating blocks add to the digit sum · Sensim Math

Variant 1 medium

$a$ and $b$ are prime numbers less than $10$ with $a > b > 0$ . The repeating decimals $0.\overline{ab}$ and $0.\overline{ba}$ (the two-digit blocks $ab$ and $ba$ repeat) add up to $0.\overline{7}$ . Write the difference of the two repeating decimals as a repeating decimal.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two repeating decimals repeat the blocks $ab$ and $ba$ . Their sum is $0.\overline{7}$ , and $a$ , $b$ are prime with $a > b > 0$ . Find their difference as a repeating decimal.

Givens

$a$ and $b$ are prime numbers less than $10$ , and $a > b > 0$ .
$0.\overline{ab} + 0.\overline{ba} = 0.\overline{7}$ .

Unknowns

$0.\overline{ab} - 0.\overline{ba}$ , written as a repeating decimal.

Constraints

$a$ and $b$ are digits from $\{2, 3, 5, 7\}$ .

2 · Planchoose the strategy

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

Write each repeating decimal as a fraction over $99$ using place value. Adding the reversed blocks collapses to a statement about $a + b$ alone; then the prime condition pins down the digits.

3 · Execute4 carry out the plan

1Write both decimals as fractions

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

A two-digit repeating block over

99

: the block

ab

is the number

10a + b

.

0.\overline{ab} = \frac{10a + b}{99}, \quad 0.\overline{ba} = \frac{10b + a}{99}

Place value turns the digits into an expression.

2Add them and solve for a + b

#13 Convert to Algebra 8.EE.C.7

The sum is

\frac{11(a + b)}{99} = \frac{a + b}{9}

, and

0.\overline{7} = \frac{7}{9}

.

\frac{(10a + b) + (10b + a)}{99} = \frac{11(a + b)}{99} = \frac{a + b}{9} = \frac{7}{9} \Rightarrow a + b = 7

Reversing the digits makes the sum depend only on

a + b

.

3Find the digits

#2 Make a Systematic List 8.EE.C.7

List prime pairs with

a > b > 0

and

a + b = 7

: only

a = 5

,

b = 2

works.

a = 5, \quad b = 2

One equation with two unknowns needs the extra conditions to finish.

4Take the difference

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

The larger decimal is

0.\overline{52}

and the smaller is

0.\overline{25}

.

\frac{52}{99} - \frac{25}{99} = \frac{27}{99} = 0.\overline{27}

The difference

\frac{27}{99}

has block

27

.

Answer:

0.\overline{27}

4 · Reviewdoes it hold up?

Check the sum: $\frac{52}{99} + \frac{25}{99} = \frac{77}{99} = \frac{7}{9} = 0.\overline{7}$ . The difference $0.2727\ldots$ is also $0.5252\ldots - 0.2525\ldots$ digit by digit.

Another way: Subtract directly without fractions: $0.525252\ldots - 0.252525\ldots = 0.272727\ldots$ , since each pair of places gives $52 - 25 = 27$ with no borrowing between pairs.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Converting the repeating decimals to fractions over $99$ and back.
8.EE.C.7 Solve linear equations in one variable — Solving $\frac{a + b}{9} = \frac{7}{9}$ and using the conditions to fix $a$ and $b$.

💡Takeaway. Swap the digits and add -- place value leaves only the digit sum behind.

Variant 2 medium

$a$ and $b$ are even numbers less than $10$ with $a > b > 0$ . The repeating decimals $0.\overline{ab}$ and $0.\overline{ba}$ (the two-digit blocks $ab$ and $ba$ repeat) add up to $1.\overline{5}$ . Write the difference of the two repeating decimals as a repeating decimal.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two repeating decimals repeat the blocks $ab$ and $ba$ . Their sum is $1.\overline{5}$ , and $a$ , $b$ are even with $a > b > 0$ . Find their difference as a repeating decimal.

Givens

$a$ and $b$ are even numbers less than $10$ , and $a > b > 0$ .
$0.\overline{ab} + 0.\overline{ba} = 1.\overline{5}$ .

Unknowns

$0.\overline{ab} - 0.\overline{ba}$ , written as a repeating decimal.

Constraints

$a$ and $b$ are digits from $\{2, 4, 6, 8\}$ .

2 · Planchoose the strategy

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

Write each repeating decimal as a fraction over $99$ using place value. Adding the reversed blocks collapses to a statement about $a + b$ alone; then the evenness condition pins down the digits.

3 · Execute4 carry out the plan

1Write both decimals as fractions

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

A two-digit repeating block over

99

: the block

ab

is the number

10a + b

.

0.\overline{ab} = \frac{10a + b}{99}, \quad 0.\overline{ba} = \frac{10b + a}{99}

Place value turns the digits into an expression.

2Add them and solve for a + b

#13 Convert to Algebra 8.EE.C.7

The sum is

\frac{11(a + b)}{99} = \frac{a + b}{9}

, and

1.\overline{5} = 1 + \frac{5}{9} = \frac{14}{9}

.

\frac{(10a + b) + (10b + a)}{99} = \frac{11(a + b)}{99} = \frac{a + b}{9} = \frac{14}{9} \Rightarrow a + b = 14

Reversing the digits makes the sum depend only on

a + b

.

3Find the digits

#2 Make a Systematic List 8.EE.C.7

List even pairs with

a > b > 0

and

a + b = 14

: only

a = 8

,

b = 6

works.

a = 8, \quad b = 6

One equation with two unknowns needs the extra conditions to finish.

4Take the difference

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

The larger decimal is

0.\overline{86}

and the smaller is

0.\overline{68}

.

\frac{86}{99} - \frac{68}{99} = \frac{18}{99} = 0.\overline{18}

The difference

\frac{18}{99}

has block

18

.

Answer:

0.\overline{18}

4 · Reviewdoes it hold up?

Check the sum: $\frac{86}{99} + \frac{68}{99} = \frac{154}{99} = 1\frac{5}{9} = 1.\overline{5}$ . The difference $0.1818\ldots$ is also $0.8686\ldots - 0.6868\ldots$ digit by digit.

Another way: Subtract directly without fractions: $0.868686\ldots - 0.686868\ldots = 0.181818\ldots$ , since each pair of places gives $86 - 68 = 18$ with no borrowing between pairs.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Converting the repeating decimals to fractions over $99$ and back.
8.EE.C.7 Solve linear equations in one variable — Solving $\frac{a + b}{9} = \frac{14}{9}$ and using the conditions to fix $a$ and $b$.

💡Takeaway. Swap the digits and add -- place value leaves only the digit sum behind.

Variant 3 medium

$a$ and $b$ are even numbers less than $10$ with $a > b > 0$ . The repeating decimals $0.\overline{ab}$ and $0.\overline{ba}$ (the two-digit blocks $ab$ and $ba$ repeat) add up to $0.\overline{8}$ . Write the difference of the two repeating decimals as a repeating decimal.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two repeating decimals repeat the blocks $ab$ and $ba$ . Their sum is $0.\overline{8}$ , and $a$ , $b$ are even with $a > b > 0$ . Find their difference as a repeating decimal.

Givens

$a$ and $b$ are even numbers less than $10$ , and $a > b > 0$ .
$0.\overline{ab} + 0.\overline{ba} = 0.\overline{8}$ .

Unknowns

$0.\overline{ab} - 0.\overline{ba}$ , written as a repeating decimal.

Constraints

$a$ and $b$ are digits from $\{2, 4, 6, 8\}$ .

2 · Planchoose the strategy

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

Write each repeating decimal as a fraction over $99$ using place value. Adding the reversed blocks collapses to a statement about $a + b$ alone; then the evenness condition pins down the digits.

3 · Execute4 carry out the plan

1Write both decimals as fractions

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

A two-digit repeating block over

99

: the block

ab

is the number

10a + b

.

0.\overline{ab} = \frac{10a + b}{99}, \quad 0.\overline{ba} = \frac{10b + a}{99}

Place value turns the digits into an expression.

2Add them and solve for a + b

#13 Convert to Algebra 8.EE.C.7

The sum is

\frac{11(a + b)}{99} = \frac{a + b}{9}

, and

0.\overline{8} = \frac{8}{9}

.

\frac{(10a + b) + (10b + a)}{99} = \frac{11(a + b)}{99} = \frac{a + b}{9} = \frac{8}{9} \Rightarrow a + b = 8

Reversing the digits makes the sum depend only on

a + b

.

3Find the digits

#2 Make a Systematic List 8.EE.C.7

List even pairs with

a > b > 0

and

a + b = 8

: only

a = 6

,

b = 2

works (

8 + 0

is ruled out because

b > 0

;

4 + 4

is ruled out because

a > b

).

a = 6, \quad b = 2

One equation with two unknowns needs the extra conditions to finish.

4Take the difference

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

The larger decimal is

0.\overline{62}

and the smaller is

0.\overline{26}

.

\frac{62}{99} - \frac{26}{99} = \frac{36}{99} = 0.\overline{36}

The difference

\frac{36}{99}

has block

36

.

Answer:

0.\overline{36}

4 · Reviewdoes it hold up?

Check the sum: $\frac{62}{99} + \frac{26}{99} = \frac{88}{99} = \frac{8}{9} = 0.\overline{8}$ . The difference $0.3636\ldots$ is also $0.6262\ldots - 0.2626\ldots$ digit by digit.

Another way: Subtract directly without fractions: $0.626262\ldots - 0.262626\ldots = 0.363636\ldots$ , since each pair of places gives $62 - 26 = 36$ with no borrowing between pairs.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Converting the repeating decimals to fractions over $99$ and back.
8.EE.C.7 Solve linear equations in one variable — Solving $\frac{a + b}{9} = \frac{8}{9}$ and using the conditions to fix $a$ and $b$.

💡Takeaway. Swap the digits and add -- place value leaves only the digit sum behind.

Variant 4 medium

$a$ and $b$ are odd numbers less than $10$ with $a > b > 0$ . The repeating decimals $0.\overline{ab}$ and $0.\overline{ba}$ (the two-digit blocks $ab$ and $ba$ repeat) add up to $1.\overline{7}$ . Write the difference of the two repeating decimals as a repeating decimal.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two repeating decimals repeat the blocks $ab$ and $ba$ . Their sum is $1.\overline{7}$ , and $a$ , $b$ are odd with $a > b > 0$ . Find their difference as a repeating decimal.

Givens

$a$ and $b$ are odd numbers less than $10$ , and $a > b > 0$ .
$0.\overline{ab} + 0.\overline{ba} = 1.\overline{7}$ .

Unknowns

$0.\overline{ab} - 0.\overline{ba}$ , written as a repeating decimal.

Constraints

$a$ and $b$ are digits from $\{1, 3, 5, 7, 9\}$ .

2 · Planchoose the strategy

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

Write each repeating decimal as a fraction over $99$ using place value. Adding the reversed blocks collapses to a statement about $a + b$ alone; then the oddness condition pins down the digits.

3 · Execute4 carry out the plan

1Write both decimals as fractions

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

A two-digit repeating block over

99

: the block

ab

is the number

10a + b

.

0.\overline{ab} = \frac{10a + b}{99}, \quad 0.\overline{ba} = \frac{10b + a}{99}

Place value turns the digits into an expression.

2Add them and solve for a + b

#13 Convert to Algebra 8.EE.C.7

The sum is

\frac{11(a + b)}{99} = \frac{a + b}{9}

, and

1.\overline{7} = 1 + \frac{7}{9} = \frac{16}{9}

.

\frac{(10a + b) + (10b + a)}{99} = \frac{11(a + b)}{99} = \frac{a + b}{9} = \frac{16}{9} \Rightarrow a + b = 16

Reversing the digits makes the sum depend only on

a + b

.

3Find the digits

#2 Make a Systematic List 8.EE.C.7

List odd pairs with

a > b > 0

and

a + b = 16

: only

a = 9

,

b = 7

works.

a = 9, \quad b = 7

One equation with two unknowns needs the extra conditions to finish.

4Take the difference

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

The larger decimal is

0.\overline{97}

and the smaller is

0.\overline{79}

.

\frac{97}{99} - \frac{79}{99} = \frac{18}{99} = 0.\overline{18}

The difference

\frac{18}{99}

has block

18

.

Answer:

0.\overline{18}

4 · Reviewdoes it hold up?

Check the sum: $\frac{97}{99} + \frac{79}{99} = \frac{176}{99} = 1\frac{7}{9} = 1.\overline{7}$ . The difference $0.1818\ldots$ is also $0.9797\ldots - 0.7979\ldots$ digit by digit.

Another way: Subtract directly without fractions: $0.979797\ldots - 0.797979\ldots = 0.181818\ldots$ , since each pair of places gives $97 - 79 = 18$ with no borrowing between pairs.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Converting the repeating decimals to fractions over $99$ and back.
8.EE.C.7 Solve linear equations in one variable — Solving $\frac{a + b}{9} = \frac{16}{9}$ and using the conditions to fix $a$ and $b$.

💡Takeaway. Swap the digits and add -- place value leaves only the digit sum behind.

Variant 5 medium

$a$ and $b$ are odd numbers less than $10$ with $a > b > 0$ . The repeating decimals $0.\overline{ab}$ and $0.\overline{ba}$ (the two-digit blocks $ab$ and $ba$ repeat) add up to $0.\overline{6}$ . Write the difference of the two repeating decimals as a repeating decimal.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two repeating decimals repeat the blocks $ab$ and $ba$ . Their sum is $0.\overline{6}$ , and $a$ , $b$ are odd with $a > b > 0$ . Find their difference as a repeating decimal.

Givens

$a$ and $b$ are odd numbers less than $10$ , and $a > b > 0$ .
$0.\overline{ab} + 0.\overline{ba} = 0.\overline{6}$ .

Unknowns

$0.\overline{ab} - 0.\overline{ba}$ , written as a repeating decimal.

Constraints

$a$ and $b$ are digits from $\{1, 3, 5, 7, 9\}$ .

2 · Planchoose the strategy

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

Write each repeating decimal as a fraction over $99$ using place value. Adding the reversed blocks collapses to a statement about $a + b$ alone; then the oddness condition pins down the digits.

3 · Execute4 carry out the plan

1Write both decimals as fractions

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

A two-digit repeating block over

99

: the block

ab

is the number

10a + b

.

0.\overline{ab} = \frac{10a + b}{99}, \quad 0.\overline{ba} = \frac{10b + a}{99}

Place value turns the digits into an expression.

2Add them and solve for a + b

#13 Convert to Algebra 8.EE.C.7

The sum is

\frac{11(a + b)}{99} = \frac{a + b}{9}

, and

0.\overline{6} = \frac{6}{9}

.

\frac{(10a + b) + (10b + a)}{99} = \frac{11(a + b)}{99} = \frac{a + b}{9} = \frac{6}{9} \Rightarrow a + b = 6

Reversing the digits makes the sum depend only on

a + b

.

3Find the digits

#2 Make a Systematic List 8.EE.C.7

List odd pairs with

a > b > 0

and

a + b = 6

: only

a = 5

,

b = 1

works (

3 + 3

is ruled out because

a > b

).

a = 5, \quad b = 1

One equation with two unknowns needs the extra conditions to finish.

4Take the difference

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

The larger decimal is

0.\overline{51}

and the smaller is

0.\overline{15}

.

\frac{51}{99} - \frac{15}{99} = \frac{36}{99} = 0.\overline{36}

The difference

\frac{36}{99}

has block

36

.

Answer:

0.\overline{36}

4 · Reviewdoes it hold up?

Check the sum: $\frac{51}{99} + \frac{15}{99} = \frac{66}{99} = \frac{2}{3} = 0.\overline{6}$ . The difference $0.3636\ldots$ is also $0.5151\ldots - 0.1515\ldots$ digit by digit.

Another way: Subtract directly without fractions: $0.515151\ldots - 0.151515\ldots = 0.363636\ldots$ , since each pair of places gives $51 - 15 = 36$ with no borrowing between pairs.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Converting the repeating decimals to fractions over $99$ and back.
8.EE.C.7 Solve linear equations in one variable — Solving $\frac{a + b}{9} = \frac{6}{9}$ and using the conditions to fix $a$ and $b$.

💡Takeaway. Swap the digits and add -- place value leaves only the digit sum behind.

Variant 6 medium

$a$ and $b$ are even numbers less than $10$ with $a > b > 0$ . The repeating decimals $0.\overline{ab}$ and $0.\overline{ba}$ (the two-digit blocks $ab$ and $ba$ repeat) add up to $0.\overline{6}$ . Write the difference of the two repeating decimals as a repeating decimal.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two repeating decimals repeat the blocks $ab$ and $ba$ . Their sum is $0.\overline{6}$ , and $a$ , $b$ are even with $a > b > 0$ . Find their difference as a repeating decimal.

Givens

$a$ and $b$ are even numbers less than $10$ , and $a > b > 0$ .
$0.\overline{ab} + 0.\overline{ba} = 0.\overline{6}$ .

Unknowns

$0.\overline{ab} - 0.\overline{ba}$ , written as a repeating decimal.

Constraints

$a$ and $b$ are digits from $\{2, 4, 6, 8\}$ .

2 · Planchoose the strategy

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

Write each repeating decimal as a fraction over $99$ using place value. Adding the reversed blocks collapses to a statement about $a + b$ alone; then the evenness condition pins down the digits.

3 · Execute4 carry out the plan

1Write both decimals as fractions

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

A two-digit repeating block over

99

: the block

ab

is the number

10a + b

.

0.\overline{ab} = \frac{10a + b}{99}, \quad 0.\overline{ba} = \frac{10b + a}{99}

Place value turns the digits into an expression.

2Add them and solve for a + b

#13 Convert to Algebra 8.EE.C.7

The sum is

\frac{11(a + b)}{99} = \frac{a + b}{9}

, and

0.\overline{6} = \frac{6}{9}

.

\frac{(10a + b) + (10b + a)}{99} = \frac{11(a + b)}{99} = \frac{a + b}{9} = \frac{6}{9} \Rightarrow a + b = 6

Reversing the digits makes the sum depend only on

a + b

.

3Find the digits

#2 Make a Systematic List 8.EE.C.7

List even pairs with

a > b > 0

and

a + b = 6

: only

a = 4

,

b = 2

works (

6 + 0

is ruled out because

b > 0

).

a = 4, \quad b = 2

One equation with two unknowns needs the extra conditions to finish.

4Take the difference

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

The larger decimal is

0.\overline{42}

and the smaller is

0.\overline{24}

.

\frac{42}{99} - \frac{24}{99} = \frac{18}{99} = 0.\overline{18}

The difference

\frac{18}{99}

has block

18

.

Answer:

0.\overline{18}

4 · Reviewdoes it hold up?

Check the sum: $\frac{42}{99} + \frac{24}{99} = \frac{66}{99} = \frac{2}{3} = 0.\overline{6}$ . The difference $0.1818\ldots$ is also $0.4242\ldots - 0.2424\ldots$ digit by digit.

Another way: Subtract directly without fractions: $0.424242\ldots - 0.242424\ldots = 0.181818\ldots$ , since each pair of places gives $42 - 24 = 18$ with no borrowing between pairs.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Converting the repeating decimals to fractions over $99$ and back.
8.EE.C.7 Solve linear equations in one variable — Solving $\frac{a + b}{9} = \frac{6}{9}$ and using the conditions to fix $a$ and $b$.

💡Takeaway. Swap the digits and add -- place value leaves only the digit sum behind.

Variant 7 medium

$a$ and $b$ are even numbers less than $10$ with $a > b > 0$ . The repeating decimals $0.\overline{ab}$ and $0.\overline{ba}$ (the two-digit blocks $ab$ and $ba$ repeat) add up to $1.\overline{3}$ . Write the difference of the two repeating decimals as a repeating decimal.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two repeating decimals repeat the blocks $ab$ and $ba$ . Their sum is $1.\overline{3}$ , and $a$ , $b$ are even with $a > b > 0$ . Find their difference as a repeating decimal.

Givens

$a$ and $b$ are even numbers less than $10$ , and $a > b > 0$ .
$0.\overline{ab} + 0.\overline{ba} = 1.\overline{3}$ .

Unknowns

$0.\overline{ab} - 0.\overline{ba}$ , written as a repeating decimal.

Constraints

$a$ and $b$ are digits from $\{2, 4, 6, 8\}$ .

2 · Planchoose the strategy

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

Write each repeating decimal as a fraction over $99$ using place value. Adding the reversed blocks collapses to a statement about $a + b$ alone; then the evenness condition pins down the digits.

3 · Execute4 carry out the plan

1Write both decimals as fractions

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

A two-digit repeating block over

99

: the block

ab

is the number

10a + b

.

0.\overline{ab} = \frac{10a + b}{99}, \quad 0.\overline{ba} = \frac{10b + a}{99}

Place value turns the digits into an expression.

2Add them and solve for a + b

#13 Convert to Algebra 8.EE.C.7

The sum is

\frac{11(a + b)}{99} = \frac{a + b}{9}

, and

1.\overline{3} = 1 + \frac{3}{9} = \frac{12}{9}

.

\frac{(10a + b) + (10b + a)}{99} = \frac{11(a + b)}{99} = \frac{a + b}{9} = \frac{12}{9} \Rightarrow a + b = 12

Reversing the digits makes the sum depend only on

a + b

.

3Find the digits

#2 Make a Systematic List 8.EE.C.7

List even pairs with

a > b > 0

and

a + b = 12

: only

a = 8

,

b = 4

works (

6 + 6

is ruled out because

a > b

).

a = 8, \quad b = 4

One equation with two unknowns needs the extra conditions to finish.

4Take the difference

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

The larger decimal is

0.\overline{84}

and the smaller is

0.\overline{48}

.

\frac{84}{99} - \frac{48}{99} = \frac{36}{99} = 0.\overline{36}

The difference

\frac{36}{99}

has block

36

.

Answer:

0.\overline{36}

4 · Reviewdoes it hold up?

Check the sum: $\frac{84}{99} + \frac{48}{99} = \frac{132}{99} = 1\frac{1}{3} = 1.\overline{3}$ . The difference $0.3636\ldots$ is also $0.8484\ldots - 0.4848\ldots$ digit by digit.

Another way: Subtract directly without fractions: $0.848484\ldots - 0.484848\ldots = 0.363636\ldots$ , since each pair of places gives $84 - 48 = 36$ with no borrowing between pairs.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Converting the repeating decimals to fractions over $99$ and back.
8.EE.C.7 Solve linear equations in one variable — Solving $\frac{a + b}{9} = \frac{12}{9}$ and using the conditions to fix $a$ and $b$.

💡Takeaway. Swap the digits and add -- place value leaves only the digit sum behind.

Variant 8 medium

$a$ and $b$ are prime numbers less than $10$ with $a > b > 0$ . The repeating decimals $0.\overline{ab}$ and $0.\overline{ba}$ (the two-digit blocks $ab$ and $ba$ repeat) add up to $0.\overline{5}$ . Write the difference of the two repeating decimals as a repeating decimal.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two repeating decimals repeat the blocks $ab$ and $ba$ . Their sum is $0.\overline{5}$ , and $a$ , $b$ are prime with $a > b > 0$ . Find their difference as a repeating decimal.

Givens

$a$ and $b$ are prime numbers less than $10$ , and $a > b > 0$ .
$0.\overline{ab} + 0.\overline{ba} = 0.\overline{5}$ .

Unknowns

$0.\overline{ab} - 0.\overline{ba}$ , written as a repeating decimal.

Constraints

$a$ and $b$ are digits from $\{2, 3, 5, 7\}$ .

2 · Planchoose the strategy

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

Write each repeating decimal as a fraction over $99$ using place value. Adding the reversed blocks collapses to a statement about $a + b$ alone; then the prime condition pins down the digits.

3 · Execute4 carry out the plan

1Write both decimals as fractions

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

A two-digit repeating block over

99

: the block

ab

is the number

10a + b

.

0.\overline{ab} = \frac{10a + b}{99}, \quad 0.\overline{ba} = \frac{10b + a}{99}

Place value turns the digits into an expression.

2Add them and solve for a + b

#13 Convert to Algebra 8.EE.C.7

The sum is

\frac{11(a + b)}{99} = \frac{a + b}{9}

, and

0.\overline{5} = \frac{5}{9}

.

\frac{(10a + b) + (10b + a)}{99} = \frac{11(a + b)}{99} = \frac{a + b}{9} = \frac{5}{9} \Rightarrow a + b = 5

Reversing the digits makes the sum depend only on

a + b

.

3Find the digits

#2 Make a Systematic List 8.EE.C.7

List prime pairs with

a > b > 0

and

a + b = 5

: only

a = 3

,

b = 2

works.

a = 3, \quad b = 2

One equation with two unknowns needs the extra conditions to finish.

4Take the difference

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

The larger decimal is

0.\overline{32}

and the smaller is

0.\overline{23}

.

\frac{32}{99} - \frac{23}{99} = \frac{9}{99} = 0.\overline{09}

The difference

\frac{9}{99}

has block

09

.

Answer:

0.\overline{09}

4 · Reviewdoes it hold up?

Check the sum: $\frac{32}{99} + \frac{23}{99} = \frac{55}{99} = \frac{5}{9} = 0.\overline{5}$ . The difference $0.0909\ldots$ is also $0.3232\ldots - 0.2323\ldots$ digit by digit.

Another way: Subtract directly without fractions: $0.323232\ldots - 0.232323\ldots = 0.090909\ldots$ , since each pair of places gives $32 - 23 = 9$ with no borrowing between pairs.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Converting the repeating decimals to fractions over $99$ and back.
8.EE.C.7 Solve linear equations in one variable — Solving $\frac{a + b}{9} = \frac{5}{9}$ and using the conditions to fix $a$ and $b$.

💡Takeaway. Swap the digits and add -- place value leaves only the digit sum behind.

Variant 9 medium

$a$ and $b$ are prime numbers less than $10$ with $a > b > 0$ . The repeating decimals $0.\overline{ab}$ and $0.\overline{ba}$ (the two-digit blocks $ab$ and $ba$ repeat) add up to $1.\overline{1}$ . Write the difference of the two repeating decimals as a repeating decimal.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two repeating decimals repeat the blocks $ab$ and $ba$ . Their sum is $1.\overline{1}$ , and $a$ , $b$ are prime with $a > b > 0$ . Find their difference as a repeating decimal.

Givens

$a$ and $b$ are prime numbers less than $10$ , and $a > b > 0$ .
$0.\overline{ab} + 0.\overline{ba} = 1.\overline{1}$ .

Unknowns

$0.\overline{ab} - 0.\overline{ba}$ , written as a repeating decimal.

Constraints

$a$ and $b$ are digits from $\{2, 3, 5, 7\}$ .

2 · Planchoose the strategy

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

Write each repeating decimal as a fraction over $99$ using place value. Adding the reversed blocks collapses to a statement about $a + b$ alone; then the prime condition pins down the digits.

3 · Execute4 carry out the plan

1Write both decimals as fractions

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

A two-digit repeating block over

99

: the block

ab

is the number

10a + b

.

0.\overline{ab} = \frac{10a + b}{99}, \quad 0.\overline{ba} = \frac{10b + a}{99}

Place value turns the digits into an expression.

2Add them and solve for a + b

#13 Convert to Algebra 8.EE.C.7

The sum is

\frac{11(a + b)}{99} = \frac{a + b}{9}

, and

1.\overline{1} = 1 + \frac{1}{9} = \frac{10}{9}

.

\frac{(10a + b) + (10b + a)}{99} = \frac{11(a + b)}{99} = \frac{a + b}{9} = \frac{10}{9} \Rightarrow a + b = 10

Reversing the digits makes the sum depend only on

a + b

.

3Find the digits

#2 Make a Systematic List 8.EE.C.7

List prime pairs with

a > b > 0

and

a + b = 10

: only

a = 7

,

b = 3

works (

5 + 5

is ruled out because

a > b

).

a = 7, \quad b = 3

One equation with two unknowns needs the extra conditions to finish.

4Take the difference

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

The larger decimal is

0.\overline{73}

and the smaller is

0.\overline{37}

.

\frac{73}{99} - \frac{37}{99} = \frac{36}{99} = 0.\overline{36}

The difference

\frac{36}{99}

has block

36

.

Answer:

0.\overline{36}

4 · Reviewdoes it hold up?

Check the sum: $\frac{73}{99} + \frac{37}{99} = \frac{110}{99} = 1\frac{1}{9} = 1.\overline{1}$ . The difference $0.3636\ldots$ is also $0.7373\ldots - 0.3737\ldots$ digit by digit.

Another way: Subtract directly without fractions: $0.737373\ldots - 0.373737\ldots = 0.363636\ldots$ , since each pair of places gives $73 - 37 = 36$ with no borrowing between pairs.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Converting the repeating decimals to fractions over $99$ and back.
8.EE.C.7 Solve linear equations in one variable — Solving $\frac{a + b}{9} = \frac{10}{9}$ and using the conditions to fix $a$ and $b$.

💡Takeaway. Swap the digits and add -- place value leaves only the digit sum behind.

Variant 10 medium

$a$ and $b$ are odd numbers less than $10$ with $a > b > 0$ . The repeating decimals $0.\overline{ab}$ and $0.\overline{ba}$ (the two-digit blocks $ab$ and $ba$ repeat) add up to $1.\overline{5}$ . Write the difference of the two repeating decimals as a repeating decimal.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two repeating decimals repeat the blocks $ab$ and $ba$ . Their sum is $1.\overline{5}$ , and $a$ , $b$ are odd with $a > b > 0$ . Find their difference as a repeating decimal.

Givens

$a$ and $b$ are odd numbers less than $10$ , and $a > b > 0$ .
$0.\overline{ab} + 0.\overline{ba} = 1.\overline{5}$ .

Unknowns

$0.\overline{ab} - 0.\overline{ba}$ , written as a repeating decimal.

Constraints

$a$ and $b$ are digits from $\{1, 3, 5, 7, 9\}$ .

2 · Planchoose the strategy

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

Write each repeating decimal as a fraction over $99$ using place value. Adding the reversed blocks collapses to a statement about $a + b$ alone; then the oddness condition pins down the digits.

3 · Execute4 carry out the plan

1Write both decimals as fractions

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

A two-digit repeating block over

99

: the block

ab

is the number

10a + b

.

0.\overline{ab} = \frac{10a + b}{99}, \quad 0.\overline{ba} = \frac{10b + a}{99}

Place value turns the digits into an expression.

2Add them and solve for a + b

#13 Convert to Algebra 8.EE.C.7

The sum is

\frac{11(a + b)}{99} = \frac{a + b}{9}

, and

1.\overline{5} = 1 + \frac{5}{9} = \frac{14}{9}

.

\frac{(10a + b) + (10b + a)}{99} = \frac{11(a + b)}{99} = \frac{a + b}{9} = \frac{14}{9} \Rightarrow a + b = 14

Reversing the digits makes the sum depend only on

a + b

.

3Find the digits

#2 Make a Systematic List 8.EE.C.7

List odd pairs with

a > b > 0

and

a + b = 14

: only

a = 9

,

b = 5

works (

7 + 7

is ruled out because

a > b

).

a = 9, \quad b = 5

One equation with two unknowns needs the extra conditions to finish.

4Take the difference

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

The larger decimal is

0.\overline{95}

and the smaller is

0.\overline{59}

.

\frac{95}{99} - \frac{59}{99} = \frac{36}{99} = 0.\overline{36}

The difference

\frac{36}{99}

has block

36

.

Answer:

0.\overline{36}

4 · Reviewdoes it hold up?

Check the sum: $\frac{95}{99} + \frac{59}{99} = \frac{154}{99} = 1\frac{5}{9} = 1.\overline{5}$ . The difference $0.3636\ldots$ is also $0.9595\ldots - 0.5959\ldots$ digit by digit.

Another way: Subtract directly without fractions: $0.959595\ldots - 0.595959\ldots = 0.363636\ldots$ , since each pair of places gives $95 - 59 = 36$ with no borrowing between pairs.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Converting the repeating decimals to fractions over $99$ and back.
8.EE.C.7 Solve linear equations in one variable — Solving $\frac{a + b}{9} = \frac{14}{9}$ and using the conditions to fix $a$ and $b$.

💡Takeaway. Swap the digits and add -- place value leaves only the digit sum behind.

Variant 11 medium

$a$ and $b$ are prime numbers less than $10$ with $a > b > 0$ . The repeating decimals $0.\overline{ab}$ and $0.\overline{ba}$ (the two-digit blocks $ab$ and $ba$ repeat) add up to $0.\overline{8}$ . Write the difference of the two repeating decimals as a repeating decimal.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two repeating decimals repeat the blocks $ab$ and $ba$ . Their sum is $0.\overline{8}$ , and $a$ , $b$ are prime with $a > b > 0$ . Find their difference as a repeating decimal.

Givens

$a$ and $b$ are prime numbers less than $10$ , and $a > b > 0$ .
$0.\overline{ab} + 0.\overline{ba} = 0.\overline{8}$ .

Unknowns

$0.\overline{ab} - 0.\overline{ba}$ , written as a repeating decimal.

Constraints

$a$ and $b$ are digits from $\{2, 3, 5, 7\}$ .

2 · Planchoose the strategy

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

Write each repeating decimal as a fraction over $99$ using place value. Adding the reversed blocks collapses to a statement about $a + b$ alone; then the prime condition pins down the digits.

3 · Execute4 carry out the plan

1Write both decimals as fractions

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

A two-digit repeating block over

99

: the block

ab

is the number

10a + b

.

0.\overline{ab} = \frac{10a + b}{99}, \quad 0.\overline{ba} = \frac{10b + a}{99}

Place value turns the digits into an expression.

2Add them and solve for a + b

#13 Convert to Algebra 8.EE.C.7

The sum is

\frac{11(a + b)}{99} = \frac{a + b}{9}

, and

0.\overline{8} = \frac{8}{9}

.

\frac{(10a + b) + (10b + a)}{99} = \frac{11(a + b)}{99} = \frac{a + b}{9} = \frac{8}{9} \Rightarrow a + b = 8

Reversing the digits makes the sum depend only on

a + b

.

3Find the digits

#2 Make a Systematic List 8.EE.C.7

List prime pairs with

a > b > 0

and

a + b = 8

: only

a = 5

,

b = 3

works.

a = 5, \quad b = 3

One equation with two unknowns needs the extra conditions to finish.

4Take the difference

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

The larger decimal is

0.\overline{53}

and the smaller is

0.\overline{35}

.

\frac{53}{99} - \frac{35}{99} = \frac{18}{99} = 0.\overline{18}

The difference

\frac{18}{99}

has block

18

.

Answer:

0.\overline{18}

4 · Reviewdoes it hold up?

Check the sum: $\frac{53}{99} + \frac{35}{99} = \frac{88}{99} = \frac{8}{9} = 0.\overline{8}$ . The difference $0.1818\ldots$ is also $0.5353\ldots - 0.3535\ldots$ digit by digit.

Another way: Subtract directly without fractions: $0.535353\ldots - 0.353535\ldots = 0.181818\ldots$ , since each pair of places gives $53 - 35 = 18$ with no borrowing between pairs.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Converting the repeating decimals to fractions over $99$ and back.
8.EE.C.7 Solve linear equations in one variable — Solving $\frac{a + b}{9} = \frac{8}{9}$ and using the conditions to fix $a$ and $b$.

💡Takeaway. Swap the digits and add -- place value leaves only the digit sum behind.

Variant 12 medium

$a$ and $b$ are odd numbers less than $10$ with $a > b > 0$ . The repeating decimals $0.\overline{ab}$ and $0.\overline{ba}$ (the two-digit blocks $ab$ and $ba$ repeat) add up to $0.\overline{4}$ . Write the difference of the two repeating decimals as a repeating decimal.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two repeating decimals repeat the blocks $ab$ and $ba$ . Their sum is $0.\overline{4}$ , and $a$ , $b$ are odd with $a > b > 0$ . Find their difference as a repeating decimal.

Givens

$a$ and $b$ are odd numbers less than $10$ , and $a > b > 0$ .
$0.\overline{ab} + 0.\overline{ba} = 0.\overline{4}$ .

Unknowns

$0.\overline{ab} - 0.\overline{ba}$ , written as a repeating decimal.

Constraints

$a$ and $b$ are digits from $\{1, 3, 5, 7, 9\}$ .

2 · Planchoose the strategy

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

Write each repeating decimal as a fraction over $99$ using place value. Adding the reversed blocks collapses to a statement about $a + b$ alone; then the oddness condition pins down the digits.

3 · Execute4 carry out the plan

1Write both decimals as fractions

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

A two-digit repeating block over

99

: the block

ab

is the number

10a + b

.

0.\overline{ab} = \frac{10a + b}{99}, \quad 0.\overline{ba} = \frac{10b + a}{99}

Place value turns the digits into an expression.

2Add them and solve for a + b

#13 Convert to Algebra 8.EE.C.7

The sum is

\frac{11(a + b)}{99} = \frac{a + b}{9}

, and

0.\overline{4} = \frac{4}{9}

.

\frac{(10a + b) + (10b + a)}{99} = \frac{11(a + b)}{99} = \frac{a + b}{9} = \frac{4}{9} \Rightarrow a + b = 4

Reversing the digits makes the sum depend only on

a + b

.

3Find the digits

#2 Make a Systematic List 8.EE.C.7

List odd pairs with

a > b > 0

and

a + b = 4

: only

a = 3

,

b = 1

works.

a = 3, \quad b = 1

One equation with two unknowns needs the extra conditions to finish.

4Take the difference

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

The larger decimal is

0.\overline{31}

and the smaller is

0.\overline{13}

.

\frac{31}{99} - \frac{13}{99} = \frac{18}{99} = 0.\overline{18}

The difference

\frac{18}{99}

has block

18

.

Answer:

0.\overline{18}

4 · Reviewdoes it hold up?

Check the sum: $\frac{31}{99} + \frac{13}{99} = \frac{44}{99} = \frac{4}{9} = 0.\overline{4}$ . The difference $0.1818\ldots$ is also $0.3131\ldots - 0.1313\ldots$ digit by digit.

Another way: Subtract directly without fractions: $0.313131\ldots - 0.131313\ldots = 0.181818\ldots$ , since each pair of places gives $31 - 13 = 18$ with no borrowing between pairs.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Converting the repeating decimals to fractions over $99$ and back.
8.EE.C.7 Solve linear equations in one variable — Solving $\frac{a + b}{9} = \frac{4}{9}$ and using the conditions to fix $a$ and $b$.

💡Takeaway. Swap the digits and add -- place value leaves only the digit sum behind.

Reversed repeating blocks add to the digit sum · 12 practice problems

Generated variants — 12

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Unknowns

Constraints

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Constraints

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Constraints

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Unknowns

Constraints

Givens

Unknowns

Constraints

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Givens

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Where next?