Practice · Too few equations: let digit limits decide · Sensim Math

Variant 1 easy

$a$ and $b$ are two-digit natural numbers. $a$ is a multiple of $3$ and is $33$ greater than $b$ . Also, the number formed by swapping the ones digit and the tens digit of $b$ is $6$ less than $a$ . Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two two-digit numbers satisfy: a = b + 33, a is a multiple of 3, and reversing b's digits gives a - 6. Find a and b.

Givens

$a$ and $b$ are two-digit natural numbers.
$a = b + 33$ .
$a$ is a multiple of $3$ .
Reversing the digits of $b$ gives a number $6$ less than $a$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Digits are whole numbers from $0$ to $9$ ; a two-digit number lies between $10$ and $99$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Writing $b$ by its digits turns the two conditions into one relation between the digits. That leaves several candidates, so list them using the two-digit limits and keep the one that makes $a$ a multiple of $3$ .

3 · Execute4 carry out the plan

1Write b by its digits

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

Let

b

have tens digit

m

and ones digit

n

, so

b = 10m + n

and its reversal is

10n + m

.

b = 10m + n,\quad a = 10m + n + 33

Place value turns a digit swap into algebra.

2Use the reversal condition

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

The reversal is

a - 6

, which is

b + 33 - 6 = b + 27

.

10n + m = 10m + n + 27,\quad 9n - 9m = 27,\quad n = m + 3

Swapping digits always changes a number by a multiple of

9

.

3List what the digit limits allow

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

Since

a = b + 33 \le 99

, we need

b \le 66

, and

b \ge 10

. The ones digit

n = m + 3

can be at most

9

. With

n = m + 3

,

b = 10m + m + 3

.

b = 14,\ 25,\ 36,\ 47,\ 58 \quad\Rightarrow\quad a = 47,\ 58,\ 69,\ 80,\ 91

The next one,

b = 69

, would make

a = 102

, which has three digits.

4Keep the multiple of 3

#3 Eliminate Possibilities 4.OA.B.4

The digit sums of the candidates for

a

are

11

,

13

,

15

,

8

, and

10

. Only

69

has a digit sum divisible by

3

.

69 = 3 \times 23

One extra condition picks one candidate out of five.

Answer:

a = 69

,

b = 36

4 · Reviewdoes it hold up?

$69 - 36 = 33$ ; reversing $36$ gives $63$ , and $69 - 6 = 63$ ; $69$ is a multiple of $3$ ; both are two-digit numbers.

Another way: Work with $a$ directly: $a = 11m + 36$ for $m = 1$ to $5$ . Since $36$ is a multiple of $3$ , $a$ is a multiple of $3$ exactly when $11m$ is, which forces $m = 3$ , giving $a = 69$ and $b = 36$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Combining the two conditions on $a$ and $b$ into one relation between digits.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Picking out the candidate that is a multiple of $3$.

💡Takeaway. When equations run out, the hidden limits (digits are

0

to

9

, numbers have two digits) finish the job.

Variant 2 easy

$a$ and $b$ are two-digit natural numbers. $a$ is a multiple of $3$ and is $35$ greater than $b$ . Also, the number formed by swapping the ones digit and the tens digit of $b$ is $26$ less than $a$ . Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two two-digit numbers satisfy: a = b + 35, a is a multiple of 3, and reversing b's digits gives a - 26. Find a and b.

Givens

$a$ and $b$ are two-digit natural numbers.
$a = b + 35$ .
$a$ is a multiple of $3$ .
Reversing the digits of $b$ gives a number $26$ less than $a$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Digits are whole numbers from $0$ to $9$ ; a two-digit number lies between $10$ and $99$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Writing $b$ by its digits turns the two conditions into one relation between the digits. That leaves several candidates, so list them using the two-digit limits and keep the one that makes $a$ a multiple of $3$ .

3 · Execute4 carry out the plan

1Write b by its digits

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

Let

b

have tens digit

m

and ones digit

n

, so

b = 10m + n

and its reversal is

10n + m

.

b = 10m + n,\quad a = 10m + n + 35

Place value turns a digit swap into algebra.

2Use the reversal condition

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

The reversal is

a - 26

, which is

b + 35 - 26 = b + 9

.

10n + m = 10m + n + 9,\quad 9n - 9m = 9,\quad n = m + 1

Swapping digits always changes a number by a multiple of

9

.

3List what the digit limits allow

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

Since

a = b + 35 \le 99

, we need

b \le 64

, and

b \ge 10

. The ones digit

n = m + 1

can be at most

9

. With

n = m + 1

,

b = 10m + m + 1

.

b = 12,\ 23,\ 34,\ 45,\ 56 \quad\Rightarrow\quad a = 47,\ 58,\ 69,\ 80,\ 91

The next one,

b = 67

, would make

a = 102

, which has three digits.

4Keep the multiple of 3

#3 Eliminate Possibilities 4.OA.B.4

The digit sums of the candidates for

a

are

11

,

13

,

15

,

8

, and

10

. Only

69

has a digit sum divisible by

3

.

69 = 3 \times 23

One extra condition picks one candidate out of five.

Answer:

a = 69

,

b = 34

4 · Reviewdoes it hold up?

$69 - 34 = 35$ ; reversing $34$ gives $43$ , and $69 - 26 = 43$ ; $69$ is a multiple of $3$ ; both are two-digit numbers.

Another way: Work with $a$ directly: $a = 11m + 36$ for $m = 1$ to $5$ . Since $36$ is a multiple of $3$ , $a$ is a multiple of $3$ exactly when $11m$ is, which forces $m = 3$ , giving $a = 69$ and $b = 34$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Combining the two conditions on $a$ and $b$ into one relation between digits.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Picking out the candidate that is a multiple of $3$.

💡Takeaway. When equations run out, the hidden limits (digits are

0

to

9

, numbers have two digits) finish the job.

Variant 3 easy

$a$ and $b$ are two-digit natural numbers. $a$ is a multiple of $3$ and is $37$ greater than $b$ . Also, the number formed by swapping the ones digit and the tens digit of $b$ is $19$ less than $a$ . Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two two-digit numbers satisfy: a = b + 37, a is a multiple of 3, and reversing b's digits gives a - 19. Find a and b.

Givens

$a$ and $b$ are two-digit natural numbers.
$a = b + 37$ .
$a$ is a multiple of $3$ .
Reversing the digits of $b$ gives a number $19$ less than $a$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Digits are whole numbers from $0$ to $9$ ; a two-digit number lies between $10$ and $99$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Writing $b$ by its digits turns the two conditions into one relation between the digits. That leaves several candidates, so list them using the two-digit limits and keep the one that makes $a$ a multiple of $3$ .

3 · Execute4 carry out the plan

1Write b by its digits

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

Let

b

have tens digit

m

and ones digit

n

, so

b = 10m + n

and its reversal is

10n + m

.

b = 10m + n,\quad a = 10m + n + 37

Place value turns a digit swap into algebra.

2Use the reversal condition

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

The reversal is

a - 19

, which is

b + 37 - 19 = b + 18

.

10n + m = 10m + n + 18,\quad 9n - 9m = 18,\quad n = m + 2

Swapping digits always changes a number by a multiple of

9

.

3List what the digit limits allow

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

Since

a = b + 37 \le 99

, we need

b \le 62

, and

b \ge 10

. The ones digit

n = m + 2

can be at most

9

. With

n = m + 2

,

b = 10m + m + 2

.

b = 13,\ 24,\ 35,\ 46,\ 57 \quad\Rightarrow\quad a = 50,\ 61,\ 72,\ 83,\ 94

The next one,

b = 68

, would make

a = 105

, which has three digits.

4Keep the multiple of 3

#3 Eliminate Possibilities 4.OA.B.4

The digit sums of the candidates for

a

are

5

,

7

,

9

,

11

, and

13

. Only

72

has a digit sum divisible by

3

.

72 = 3 \times 24

One extra condition picks one candidate out of five.

Answer:

a = 72

,

b = 35

4 · Reviewdoes it hold up?

$72 - 35 = 37$ ; reversing $35$ gives $53$ , and $72 - 19 = 53$ ; $72$ is a multiple of $3$ ; both are two-digit numbers.

Another way: Work with $a$ directly: $a = 11m + 39$ for $m = 1$ to $5$ . Since $39$ is a multiple of $3$ , $a$ is a multiple of $3$ exactly when $11m$ is, which forces $m = 3$ , giving $a = 72$ and $b = 35$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Combining the two conditions on $a$ and $b$ into one relation between digits.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Picking out the candidate that is a multiple of $3$.

💡Takeaway. When equations run out, the hidden limits (digits are

0

to

9

, numbers have two digits) finish the job.

Variant 4 easy

$a$ and $b$ are two-digit natural numbers. $a$ is a multiple of $3$ and is $40$ greater than $b$ . Also, the number formed by swapping the ones digit and the tens digit of $b$ is $22$ less than $a$ . Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two two-digit numbers satisfy: a = b + 40, a is a multiple of 3, and reversing b's digits gives a - 22. Find a and b.

Givens

$a$ and $b$ are two-digit natural numbers.
$a = b + 40$ .
$a$ is a multiple of $3$ .
Reversing the digits of $b$ gives a number $22$ less than $a$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Digits are whole numbers from $0$ to $9$ ; a two-digit number lies between $10$ and $99$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Writing $b$ by its digits turns the two conditions into one relation between the digits. That leaves several candidates, so list them using the two-digit limits and keep the one that makes $a$ a multiple of $3$ .

3 · Execute4 carry out the plan

1Write b by its digits

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

Let

b

have tens digit

m

and ones digit

n

, so

b = 10m + n

and its reversal is

10n + m

.

b = 10m + n,\quad a = 10m + n + 40

Place value turns a digit swap into algebra.

2Use the reversal condition

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

The reversal is

a - 22

, which is

b + 40 - 22 = b + 18

.

10n + m = 10m + n + 18,\quad 9n - 9m = 18,\quad n = m + 2

Swapping digits always changes a number by a multiple of

9

.

3List what the digit limits allow

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

Since

a = b + 40 \le 99

, we need

b \le 59

, and

b \ge 10

. The ones digit

n = m + 2

can be at most

9

. With

n = m + 2

,

b = 10m + m + 2

.

b = 13,\ 24,\ 35,\ 46,\ 57 \quad\Rightarrow\quad a = 53,\ 64,\ 75,\ 86,\ 97

The next one,

b = 68

, would make

a = 108

, which has three digits.

4Keep the multiple of 3

#3 Eliminate Possibilities 4.OA.B.4

The digit sums of the candidates for

a

are

8

,

10

,

12

,

14

, and

16

. Only

75

has a digit sum divisible by

3

.

75 = 3 \times 25

One extra condition picks one candidate out of five.

Answer:

a = 75

,

b = 35

4 · Reviewdoes it hold up?

$75 - 35 = 40$ ; reversing $35$ gives $53$ , and $75 - 22 = 53$ ; $75$ is a multiple of $3$ ; both are two-digit numbers.

Another way: Work with $a$ directly: $a = 11m + 42$ for $m = 1$ to $5$ . Since $42$ is a multiple of $3$ , $a$ is a multiple of $3$ exactly when $11m$ is, which forces $m = 3$ , giving $a = 75$ and $b = 35$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Combining the two conditions on $a$ and $b$ into one relation between digits.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Picking out the candidate that is a multiple of $3$.

💡Takeaway. When equations run out, the hidden limits (digits are

0

to

9

, numbers have two digits) finish the job.

Variant 5 medium

$a$ and $b$ are two-digit natural numbers. $a$ is a multiple of $3$ and is $44$ greater than $b$ . Also, the number formed by swapping the ones digit and the tens digit of $b$ is $8$ less than $a$ . Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two two-digit numbers satisfy: a = b + 44, a is a multiple of 3, and reversing b's digits gives a - 8. Find a and b.

Givens

$a$ and $b$ are two-digit natural numbers.
$a = b + 44$ .
$a$ is a multiple of $3$ .
Reversing the digits of $b$ gives a number $8$ less than $a$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Digits are whole numbers from $0$ to $9$ ; a two-digit number lies between $10$ and $99$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Writing $b$ by its digits turns the two conditions into one relation between the digits. That leaves several candidates, so list them using the two-digit limits and keep the one that makes $a$ a multiple of $3$ .

3 · Execute4 carry out the plan

1Write b by its digits

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

Let

b

have tens digit

m

and ones digit

n

, so

b = 10m + n

and its reversal is

10n + m

.

b = 10m + n,\quad a = 10m + n + 44

Place value turns a digit swap into algebra.

2Use the reversal condition

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

The reversal is

a - 8

, which is

b + 44 - 8 = b + 36

.

10n + m = 10m + n + 36,\quad 9n - 9m = 36,\quad n = m + 4

Swapping digits always changes a number by a multiple of

9

.

3List what the digit limits allow

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

Since

a = b + 44 \le 99

, we need

b \le 55

, and

b \ge 10

. The ones digit

n = m + 4

can be at most

9

. With

n = m + 4

,

b = 10m + m + 4

.

b = 15,\ 26,\ 37,\ 48 \quad\Rightarrow\quad a = 59,\ 70,\ 81,\ 92

The next one,

b = 59

, would make

a = 103

, which has three digits.

4Keep the multiple of 3

#3 Eliminate Possibilities 4.OA.B.4

The digit sums of the candidates for

a

are

14

,

7

,

9

, and

11

. Only

81

has a digit sum divisible by

3

.

81 = 3 \times 27

One extra condition picks one candidate out of four.

Answer:

a = 81

,

b = 37

4 · Reviewdoes it hold up?

$81 - 37 = 44$ ; reversing $37$ gives $73$ , and $81 - 8 = 73$ ; $81$ is a multiple of $3$ ; both are two-digit numbers.

Another way: Work with $a$ directly: $a = 11m + 48$ for $m = 1$ to $4$ . Since $48$ is a multiple of $3$ , $a$ is a multiple of $3$ exactly when $11m$ is, which forces $m = 3$ , giving $a = 81$ and $b = 37$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Combining the two conditions on $a$ and $b$ into one relation between digits.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Picking out the candidate that is a multiple of $3$.

💡Takeaway. When equations run out, the hidden limits (digits are

0

to

9

, numbers have two digits) finish the job.

Variant 6 medium

$a$ and $b$ are two-digit natural numbers. $a$ is a multiple of $3$ and is $46$ greater than $b$ . Also, the number formed by swapping the ones digit and the tens digit of $b$ is $28$ less than $a$ . Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two two-digit numbers satisfy: a = b + 46, a is a multiple of 3, and reversing b's digits gives a - 28. Find a and b.

Givens

$a$ and $b$ are two-digit natural numbers.
$a = b + 46$ .
$a$ is a multiple of $3$ .
Reversing the digits of $b$ gives a number $28$ less than $a$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Digits are whole numbers from $0$ to $9$ ; a two-digit number lies between $10$ and $99$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Writing $b$ by its digits turns the two conditions into one relation between the digits. That leaves several candidates, so list them using the two-digit limits and keep the one that makes $a$ a multiple of $3$ .

3 · Execute4 carry out the plan

1Write b by its digits

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

Let

b

have tens digit

m

and ones digit

n

, so

b = 10m + n

and its reversal is

10n + m

.

b = 10m + n,\quad a = 10m + n + 46

Place value turns a digit swap into algebra.

2Use the reversal condition

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

The reversal is

a - 28

, which is

b + 46 - 28 = b + 18

.

10n + m = 10m + n + 18,\quad 9n - 9m = 18,\quad n = m + 2

Swapping digits always changes a number by a multiple of

9

.

3List what the digit limits allow

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

Since

a = b + 46 \le 99

, we need

b \le 53

, and

b \ge 10

. The ones digit

n = m + 2

can be at most

9

. With

n = m + 2

,

b = 10m + m + 2

.

b = 13,\ 24,\ 35,\ 46 \quad\Rightarrow\quad a = 59,\ 70,\ 81,\ 92

The next one,

b = 57

, would make

a = 103

, which has three digits.

4Keep the multiple of 3

#3 Eliminate Possibilities 4.OA.B.4

The digit sums of the candidates for

a

are

14

,

7

,

9

, and

11

. Only

81

has a digit sum divisible by

3

.

81 = 3 \times 27

One extra condition picks one candidate out of four.

Answer:

a = 81

,

b = 35

4 · Reviewdoes it hold up?

$81 - 35 = 46$ ; reversing $35$ gives $53$ , and $81 - 28 = 53$ ; $81$ is a multiple of $3$ ; both are two-digit numbers.

Another way: Work with $a$ directly: $a = 11m + 48$ for $m = 1$ to $4$ . Since $48$ is a multiple of $3$ , $a$ is a multiple of $3$ exactly when $11m$ is, which forces $m = 3$ , giving $a = 81$ and $b = 35$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Combining the two conditions on $a$ and $b$ into one relation between digits.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Picking out the candidate that is a multiple of $3$.

💡Takeaway. When equations run out, the hidden limits (digits are

0

to

9

, numbers have two digits) finish the job.

Variant 7 medium

$a$ and $b$ are two-digit natural numbers. $a$ is a multiple of $3$ and is $49$ greater than $b$ . Also, the number formed by swapping the ones digit and the tens digit of $b$ is $4$ less than $a$ . Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two two-digit numbers satisfy: a = b + 49, a is a multiple of 3, and reversing b's digits gives a - 4. Find a and b.

Givens

$a$ and $b$ are two-digit natural numbers.
$a = b + 49$ .
$a$ is a multiple of $3$ .
Reversing the digits of $b$ gives a number $4$ less than $a$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Digits are whole numbers from $0$ to $9$ ; a two-digit number lies between $10$ and $99$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Writing $b$ by its digits turns the two conditions into one relation between the digits. That leaves several candidates, so list them using the two-digit limits and keep the one that makes $a$ a multiple of $3$ .

3 · Execute4 carry out the plan

1Write b by its digits

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

Let

b

have tens digit

m

and ones digit

n

, so

b = 10m + n

and its reversal is

10n + m

.

b = 10m + n,\quad a = 10m + n + 49

Place value turns a digit swap into algebra.

2Use the reversal condition

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

The reversal is

a - 4

, which is

b + 49 - 4 = b + 45

.

10n + m = 10m + n + 45,\quad 9n - 9m = 45,\quad n = m + 5

Swapping digits always changes a number by a multiple of

9

.

3List what the digit limits allow

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

Since

a = b + 49 \le 99

, we need

b \le 50

, and

b \ge 10

. The ones digit

n = m + 5

can be at most

9

. With

n = m + 5

,

b = 10m + m + 5

.

b = 16,\ 27,\ 38,\ 49 \quad\Rightarrow\quad a = 65,\ 76,\ 87,\ 98

The next one would need a ones digit of

10

, which is not a digit.

4Keep the multiple of 3

#3 Eliminate Possibilities 4.OA.B.4

The digit sums of the candidates for

a

are

11

,

13

,

15

, and

17

. Only

87

has a digit sum divisible by

3

.

87 = 3 \times 29

One extra condition picks one candidate out of four.

Answer:

a = 87

,

b = 38

4 · Reviewdoes it hold up?

$87 - 38 = 49$ ; reversing $38$ gives $83$ , and $87 - 4 = 83$ ; $87$ is a multiple of $3$ ; both are two-digit numbers.

Another way: Work with $a$ directly: $a = 11m + 54$ for $m = 1$ to $4$ . Since $54$ is a multiple of $3$ , $a$ is a multiple of $3$ exactly when $11m$ is, which forces $m = 3$ , giving $a = 87$ and $b = 38$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Combining the two conditions on $a$ and $b$ into one relation between digits.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Picking out the candidate that is a multiple of $3$.

💡Takeaway. When equations run out, the hidden limits (digits are

0

to

9

, numbers have two digits) finish the job.

Variant 8 medium

$a$ and $b$ are two-digit natural numbers. $a$ is a multiple of $3$ and is $51$ greater than $b$ . Also, the number formed by swapping the ones digit and the tens digit of $b$ is $24$ less than $a$ . Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two two-digit numbers satisfy: a = b + 51, a is a multiple of 3, and reversing b's digits gives a - 24. Find a and b.

Givens

$a$ and $b$ are two-digit natural numbers.
$a = b + 51$ .
$a$ is a multiple of $3$ .
Reversing the digits of $b$ gives a number $24$ less than $a$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Digits are whole numbers from $0$ to $9$ ; a two-digit number lies between $10$ and $99$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Writing $b$ by its digits turns the two conditions into one relation between the digits. That leaves several candidates, so list them using the two-digit limits and keep the one that makes $a$ a multiple of $3$ .

3 · Execute4 carry out the plan

1Write b by its digits

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

Let

b

have tens digit

m

and ones digit

n

, so

b = 10m + n

and its reversal is

10n + m

.

b = 10m + n,\quad a = 10m + n + 51

Place value turns a digit swap into algebra.

2Use the reversal condition

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

The reversal is

a - 24

, which is

b + 51 - 24 = b + 27

.

10n + m = 10m + n + 27,\quad 9n - 9m = 27,\quad n = m + 3

Swapping digits always changes a number by a multiple of

9

.

3List what the digit limits allow

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

Since

a = b + 51 \le 99

, we need

b \le 48

, and

b \ge 10

. The ones digit

n = m + 3

can be at most

9

. With

n = m + 3

,

b = 10m + m + 3

.

b = 14,\ 25,\ 36,\ 47 \quad\Rightarrow\quad a = 65,\ 76,\ 87,\ 98

The next one,

b = 58

, would make

a = 109

, which has three digits.

4Keep the multiple of 3

#3 Eliminate Possibilities 4.OA.B.4

The digit sums of the candidates for

a

are

11

,

13

,

15

, and

17

. Only

87

has a digit sum divisible by

3

.

87 = 3 \times 29

One extra condition picks one candidate out of four.

Answer:

a = 87

,

b = 36

4 · Reviewdoes it hold up?

$87 - 36 = 51$ ; reversing $36$ gives $63$ , and $87 - 24 = 63$ ; $87$ is a multiple of $3$ ; both are two-digit numbers.

Another way: Work with $a$ directly: $a = 11m + 54$ for $m = 1$ to $4$ . Since $54$ is a multiple of $3$ , $a$ is a multiple of $3$ exactly when $11m$ is, which forces $m = 3$ , giving $a = 87$ and $b = 36$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Combining the two conditions on $a$ and $b$ into one relation between digits.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Picking out the candidate that is a multiple of $3$.

💡Takeaway. When equations run out, the hidden limits (digits are

0

to

9

, numbers have two digits) finish the job.

Variant 9 hard

$a$ and $b$ are two-digit natural numbers. $a$ is a multiple of $3$ and is $55$ greater than $b$ . Also, the number formed by swapping the ones digit and the tens digit of $b$ is $37$ less than $a$ . Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two two-digit numbers satisfy: a = b + 55, a is a multiple of 3, and reversing b's digits gives a - 37. Find a and b.

Givens

$a$ and $b$ are two-digit natural numbers.
$a = b + 55$ .
$a$ is a multiple of $3$ .
Reversing the digits of $b$ gives a number $37$ less than $a$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Digits are whole numbers from $0$ to $9$ ; a two-digit number lies between $10$ and $99$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Writing $b$ by its digits turns the two conditions into one relation between the digits. That leaves several candidates, so list them using the two-digit limits and keep the one that makes $a$ a multiple of $3$ .

3 · Execute4 carry out the plan

1Write b by its digits

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

Let

b

have tens digit

m

and ones digit

n

, so

b = 10m + n

and its reversal is

10n + m

.

b = 10m + n,\quad a = 10m + n + 55

Place value turns a digit swap into algebra.

2Use the reversal condition

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

The reversal is

a - 37

, which is

b + 55 - 37 = b + 18

.

10n + m = 10m + n + 18,\quad 9n - 9m = 18,\quad n = m + 2

Swapping digits always changes a number by a multiple of

9

.

3List what the digit limits allow

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

Since

a = b + 55 \le 99

, we need

b \le 44

, and

b \ge 10

. The ones digit

n = m + 2

can be at most

9

. With

n = m + 2

,

b = 10m + m + 2

.

b = 13,\ 24,\ 35 \quad\Rightarrow\quad a = 68,\ 79,\ 90

The next one,

b = 46

, would make

a = 101

, which has three digits.

4Keep the multiple of 3

#3 Eliminate Possibilities 4.OA.B.4

The digit sums of the candidates for

a

are

14

,

16

, and

9

. Only

90

has a digit sum divisible by

3

.

90 = 3 \times 30

One extra condition picks one candidate out of three.

Answer:

a = 90

,

b = 35

4 · Reviewdoes it hold up?

$90 - 35 = 55$ ; reversing $35$ gives $53$ , and $90 - 37 = 53$ ; $90$ is a multiple of $3$ ; both are two-digit numbers.

Another way: Work with $a$ directly: $a = 11m + 57$ for $m = 1$ to $3$ . Since $57$ is a multiple of $3$ , $a$ is a multiple of $3$ exactly when $11m$ is, which forces $m = 3$ , giving $a = 90$ and $b = 35$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Combining the two conditions on $a$ and $b$ into one relation between digits.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Picking out the candidate that is a multiple of $3$.

💡Takeaway. When equations run out, the hidden limits (digits are

0

to

9

, numbers have two digits) finish the job.

Variant 10 hard

$a$ and $b$ are two-digit natural numbers. $a$ is a multiple of $3$ and is $57$ greater than $b$ . Also, the number formed by swapping the ones digit and the tens digit of $b$ is $3$ less than $a$ . Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two two-digit numbers satisfy: a = b + 57, a is a multiple of 3, and reversing b's digits gives a - 3. Find a and b.

Givens

$a$ and $b$ are two-digit natural numbers.
$a = b + 57$ .
$a$ is a multiple of $3$ .
Reversing the digits of $b$ gives a number $3$ less than $a$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Digits are whole numbers from $0$ to $9$ ; a two-digit number lies between $10$ and $99$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Writing $b$ by its digits turns the two conditions into one relation between the digits. That leaves several candidates, so list them using the two-digit limits and keep the one that makes $a$ a multiple of $3$ .

3 · Execute4 carry out the plan

1Write b by its digits

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

Let

b

have tens digit

m

and ones digit

n

, so

b = 10m + n

and its reversal is

10n + m

.

b = 10m + n,\quad a = 10m + n + 57

Place value turns a digit swap into algebra.

2Use the reversal condition

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

The reversal is

a - 3

, which is

b + 57 - 3 = b + 54

.

10n + m = 10m + n + 54,\quad 9n - 9m = 54,\quad n = m + 6

Swapping digits always changes a number by a multiple of

9

.

3List what the digit limits allow

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

Since

a = b + 57 \le 99

, we need

b \le 42

, and

b \ge 10

. The ones digit

n = m + 6

can be at most

9

. With

n = m + 6

,

b = 10m + m + 6

.

b = 17,\ 28,\ 39 \quad\Rightarrow\quad a = 74,\ 85,\ 96

The next one would need a ones digit of

10

, which is not a digit.

4Keep the multiple of 3

#3 Eliminate Possibilities 4.OA.B.4

The digit sums of the candidates for

a

are

11

,

13

, and

15

. Only

96

has a digit sum divisible by

3

.

96 = 3 \times 32

One extra condition picks one candidate out of three.

Answer:

a = 96

,

b = 39

4 · Reviewdoes it hold up?

$96 - 39 = 57$ ; reversing $39$ gives $93$ , and $96 - 3 = 93$ ; $96$ is a multiple of $3$ ; both are two-digit numbers.

Another way: Work with $a$ directly: $a = 11m + 63$ for $m = 1$ to $3$ . Since $63$ is a multiple of $3$ , $a$ is a multiple of $3$ exactly when $11m$ is, which forces $m = 3$ , giving $a = 96$ and $b = 39$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Combining the two conditions on $a$ and $b$ into one relation between digits.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Picking out the candidate that is a multiple of $3$.

💡Takeaway. When equations run out, the hidden limits (digits are

0

to

9

, numbers have two digits) finish the job.

Variant 11 hard

$a$ and $b$ are two-digit natural numbers. $a$ is a multiple of $3$ and is $60$ greater than $b$ . Also, the number formed by swapping the ones digit and the tens digit of $b$ is $33$ less than $a$ . Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two two-digit numbers satisfy: a = b + 60, a is a multiple of 3, and reversing b's digits gives a - 33. Find a and b.

Givens

$a$ and $b$ are two-digit natural numbers.
$a = b + 60$ .
$a$ is a multiple of $3$ .
Reversing the digits of $b$ gives a number $33$ less than $a$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Digits are whole numbers from $0$ to $9$ ; a two-digit number lies between $10$ and $99$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Writing $b$ by its digits turns the two conditions into one relation between the digits. That leaves several candidates, so list them using the two-digit limits and keep the one that makes $a$ a multiple of $3$ .

3 · Execute4 carry out the plan

1Write b by its digits

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

Let

b

have tens digit

m

and ones digit

n

, so

b = 10m + n

and its reversal is

10n + m

.

b = 10m + n,\quad a = 10m + n + 60

Place value turns a digit swap into algebra.

2Use the reversal condition

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

The reversal is

a - 33

, which is

b + 60 - 33 = b + 27

.

10n + m = 10m + n + 27,\quad 9n - 9m = 27,\quad n = m + 3

Swapping digits always changes a number by a multiple of

9

.

3List what the digit limits allow

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

Since

a = b + 60 \le 99

, we need

b \le 39

, and

b \ge 10

. The ones digit

n = m + 3

can be at most

9

. With

n = m + 3

,

b = 10m + m + 3

.

b = 14,\ 25,\ 36 \quad\Rightarrow\quad a = 74,\ 85,\ 96

The next one,

b = 47

, would make

a = 107

, which has three digits.

4Keep the multiple of 3

#3 Eliminate Possibilities 4.OA.B.4

The digit sums of the candidates for

a

are

11

,

13

, and

15

. Only

96

has a digit sum divisible by

3

.

96 = 3 \times 32

One extra condition picks one candidate out of three.

Answer:

a = 96

,

b = 36

4 · Reviewdoes it hold up?

$96 - 36 = 60$ ; reversing $36$ gives $63$ , and $96 - 33 = 63$ ; $96$ is a multiple of $3$ ; both are two-digit numbers.

Another way: Work with $a$ directly: $a = 11m + 63$ for $m = 1$ to $3$ . Since $63$ is a multiple of $3$ , $a$ is a multiple of $3$ exactly when $11m$ is, which forces $m = 3$ , giving $a = 96$ and $b = 36$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Combining the two conditions on $a$ and $b$ into one relation between digits.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Picking out the candidate that is a multiple of $3$.

💡Takeaway. When equations run out, the hidden limits (digits are

0

to

9

, numbers have two digits) finish the job.

Variant 12 hard

$a$ and $b$ are two-digit natural numbers. $a$ is a multiple of $3$ and is $64$ greater than $b$ . Also, the number formed by swapping the ones digit and the tens digit of $b$ is $46$ less than $a$ . Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two two-digit numbers satisfy: a = b + 64, a is a multiple of 3, and reversing b's digits gives a - 46. Find a and b.

Givens

$a$ and $b$ are two-digit natural numbers.
$a = b + 64$ .
$a$ is a multiple of $3$ .
Reversing the digits of $b$ gives a number $46$ less than $a$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Digits are whole numbers from $0$ to $9$ ; a two-digit number lies between $10$ and $99$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#3 Eliminate Possibilities

Writing $b$ by its digits turns the two conditions into one relation between the digits. That leaves several candidates, so list them using the two-digit limits and keep the one that makes $a$ a multiple of $3$ .

3 · Execute4 carry out the plan

1Write b by its digits

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

Let

b

have tens digit

m

and ones digit

n

, so

b = 10m + n

and its reversal is

10n + m

.

b = 10m + n,\quad a = 10m + n + 64

Place value turns a digit swap into algebra.

2Use the reversal condition

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

The reversal is

a - 46

, which is

b + 64 - 46 = b + 18

.

10n + m = 10m + n + 18,\quad 9n - 9m = 18,\quad n = m + 2

Swapping digits always changes a number by a multiple of

9

.

3List what the digit limits allow

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

Since

a = b + 64 \le 99

, we need

b \le 35

, and

b \ge 10

. The ones digit

n = m + 2

can be at most

9

. With

n = m + 2

,

b = 10m + m + 2

.

b = 13,\ 24,\ 35 \quad\Rightarrow\quad a = 77,\ 88,\ 99

The next one,

b = 46

, would make

a = 110

, which has three digits.

4Keep the multiple of 3

#3 Eliminate Possibilities 4.OA.B.4

The digit sums of the candidates for

a

are

14

,

16

, and

18

. Only

99

has a digit sum divisible by

3

.

99 = 3 \times 33

One extra condition picks one candidate out of three.

Answer:

a = 99

,

b = 35

4 · Reviewdoes it hold up?

$99 - 35 = 64$ ; reversing $35$ gives $53$ , and $99 - 46 = 53$ ; $99$ is a multiple of $3$ ; both are two-digit numbers.

Another way: Work with $a$ directly: $a = 11m + 66$ for $m = 1$ to $3$ . Since $66$ is a multiple of $3$ , $a$ is a multiple of $3$ exactly when $11m$ is, which forces $m = 3$ , giving $a = 99$ and $b = 35$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Combining the two conditions on $a$ and $b$ into one relation between digits.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Picking out the candidate that is a multiple of $3$.

💡Takeaway. When equations run out, the hidden limits (digits are

0

to

9

, numbers have two digits) finish the job.

Too few equations: let digit limits decide · 12 practice problems

Generated variants — 12

Givens

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Constraints

Givens

Unknowns

Constraints

Givens

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Givens

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Where next?