Practice · Run a counting rule backward to every input · Sensim Math

Variant 1 easy

For two variables $x$ and $y$ , let $y$ be the number of positive even numbers less than or equal to $x$ . For the ordered pairs $(x, y)$ made this way, find the sum of all natural numbers $a$ for which $(a, 5)$ is one of the pairs.

Your answer

Show solution

1 · Understandwhat's really being asked

Each natural number $x$ is paired with how many positive even numbers are at most $x$ . Find every $x$ that gets paired with $5$ , and add them.

Givens

$y$ is the number of positive even numbers that are less than or equal to $x$ .
The pair is $(a, 5)$ with $a$ a natural number.

Unknowns

The sum of all natural numbers $a$ with $y=5$ .

Constraints

The count only goes up when $x$ reaches a new even number, so several $x$ values can share the same $y$ .

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #2 Make a Systematic List

List the count as $x$ grows and watch the pattern: $y$ climbs by $1$ at each even number and stays put on the odd number after it. The values of $x$ with $y=5$ form a short stretch, which we list and add.

3 · Execute3 carry out the plan

1List the count near $y=5$

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

The positive even numbers in order are

2, 4, 6, 8, 10, 12, \dots

Up to

9

there are four of them;

10

brings the fifth;

12

brings the sixth.

x = 9 \to y = 4, \quad x = 10 \to y = 5, \quad x = 11 \to y = 5, \quad x = 12 \to y = 6

Just count the evens up to each

x

.

2Find the stretch where $y=5$

#5 Look for a Pattern 7.EE.B.4

Exactly five evens,

2, 4, 6, 8, 10

, are at most

a

when

a

has reached

10

but not yet

12

.

10 \le a < 12

Each count is held by an even number and the odd number right after it.

3Add the possible values

#2 Make a Systematic List 7.EE.B.4

The natural numbers in that stretch are

10

and

11

.

10 + 11 = 21

Both give the same count, so both count.

Answer:

21

4 · Reviewdoes it hold up?

For $a=10$ the evens are $2, 4, 6, 8, 10$ , which is five; for $a=11$ they are the same five; $a=9$ gives four and $a=12$ gives six. So exactly $10$ and $11$ work.

Another way: Write the count as the number of multiples of $2$ up to $a$ : it equals $5$ when $2 \times 5 \le a < 2 \times 6$ , giving $a=10$ or $11$ directly.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the condition 'exactly five evens' as the inequality $10 \le a < 12$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing multiples of $2$ up to a number.

💡Takeaway. When a rule counts things, many inputs can share an output -- find the whole stretch, not just one value.

Variant 2 easy

For two variables $x$ and $y$ , let $y$ be the number of positive even numbers less than or equal to $x$ . For the ordered pairs $(x, y)$ made this way, find the sum of all natural numbers $a$ for which $(a, 7)$ is one of the pairs.

Your answer

Show solution

1 · Understandwhat's really being asked

Each natural number $x$ is paired with how many positive even numbers are at most $x$ . Find every $x$ that gets paired with $7$ , and add them.

Givens

$y$ is the number of positive even numbers that are less than or equal to $x$ .
The pair is $(a, 7)$ with $a$ a natural number.

Unknowns

The sum of all natural numbers $a$ with $y=7$ .

Constraints

The count only goes up when $x$ reaches a new even number, so several $x$ values can share the same $y$ .

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #2 Make a Systematic List

List the count as $x$ grows and watch the pattern: $y$ climbs by $1$ at each even number and stays put on the odd number after it. The values of $x$ with $y=7$ form a short stretch, which we list and add.

3 · Execute3 carry out the plan

1List the count near $y=7$

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

The positive even numbers in order are

2, 4, 6, 8, 10, 12, 14, 16, \dots

Up to

13

there are six of them;

14

brings the seventh;

16

brings the eighth.

x = 13 \to y = 6, \quad x = 14 \to y = 7, \quad x = 15 \to y = 7, \quad x = 16 \to y = 8

Just count the evens up to each

x

.

2Find the stretch where $y=7$

#5 Look for a Pattern 7.EE.B.4

Exactly seven evens,

2, 4, \dots, 14

, are at most

a

when

a

has reached

14

but not yet

16

.

14 \le a < 16

Each count is held by an even number and the odd number right after it.

3Add the possible values

#2 Make a Systematic List 7.EE.B.4

The natural numbers in that stretch are

14

and

15

.

14 + 15 = 29

Both give the same count, so both count.

Answer:

29

4 · Reviewdoes it hold up?

For $a=14$ the evens are $2, 4, \dots, 14$ , which is seven; for $a=15$ they are the same seven; $a=13$ gives six and $a=16$ gives eight. So exactly $14$ and $15$ work.

Another way: Write the count as the number of multiples of $2$ up to $a$ : it equals $7$ when $2 \times 7 \le a < 2 \times 8$ , giving $a=14$ or $15$ directly.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the condition 'exactly seven evens' as the inequality $14 \le a < 16$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing multiples of $2$ up to a number.

💡Takeaway. When a rule counts things, many inputs can share an output -- find the whole stretch, not just one value.

Variant 3 easy

For two variables $x$ and $y$ , let $y$ be the number of positive multiples of $4$ less than or equal to $x$ . For the ordered pairs $(x, y)$ made this way, find the sum of all natural numbers $a$ for which $(a, 3)$ is one of the pairs.

Your answer

Show solution

1 · Understandwhat's really being asked

Each natural number $x$ is paired with how many positive multiples of $4$ are at most $x$ . Find every $x$ that gets paired with $3$ , and add them.

Givens

$y$ is the number of positive multiples of $4$ that are less than or equal to $x$ .
The pair is $(a, 3)$ with $a$ a natural number.

Unknowns

The sum of all natural numbers $a$ with $y=3$ .

Constraints

The count only goes up when $x$ reaches a new multiple of $4$ , so several $x$ values can share the same $y$ .

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #2 Make a Systematic List

List the count as $x$ grows and watch the pattern: $y$ climbs by $1$ at each multiple of $4$ and stays put on the $3$ numbers after it. The values of $x$ with $y=3$ form a short stretch, which we list and add.

3 · Execute3 carry out the plan

1List the count near $y=3$

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

The positive multiples of

4

in order are

4, 8, 12, 16, \dots

Up to

11

there are two of them;

12

brings the third;

16

brings the fourth.

x = 11 \to y = 2, \quad x = 12 \to y = 3, \quad x = 13 \to y = 3, \quad x = 14 \to y = 3, \quad x = 15 \to y = 3, \quad x = 16 \to y = 4

Just count the multiples of

4

up to each

x

.

2Find the stretch where $y=3$

#5 Look for a Pattern 7.EE.B.4

Exactly three multiples of

4

,

4, 8, 12

, are at most

a

when

a

has reached

12

but not yet

16

.

12 \le a < 16

Each count is held by a multiple of

4

and the

3

numbers right after it.

3Add the possible values

#2 Make a Systematic List 7.EE.B.4

The natural numbers in that stretch are

12

,

13

,

14

, and

15

.

12 + 13 + 14 + 15 = 54

All

4

give the same count, so all of them count.

Answer:

54

4 · Reviewdoes it hold up?

For $a=12$ the multiples of $4$ are $4, 8, 12$ , which is three; for $a=15$ they are the same three; $a=11$ gives two and $a=16$ gives four. So exactly $12$ , $13$ , $14$ , and $15$ work.

Another way: Write the count as the number of multiples of $4$ up to $a$ : it equals $3$ when $4 \times 3 \le a < 4 \times 4$ , giving $a$ from $12$ to $15$ directly.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the condition 'exactly three multiples of $4$' as the inequality $12 \le a < 16$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing multiples of $4$ up to a number.

💡Takeaway. When a rule counts things, many inputs can share an output -- find the whole stretch, not just one value.

Variant 4 easy

For two variables $x$ and $y$ , let $y$ be the number of positive multiples of $3$ less than or equal to $x$ . For the ordered pairs $(x, y)$ made this way, find the sum of all natural numbers $a$ for which $(a, 4)$ is one of the pairs.

Your answer

Show solution

1 · Understandwhat's really being asked

Each natural number $x$ is paired with how many positive multiples of $3$ are at most $x$ . Find every $x$ that gets paired with $4$ , and add them.

Givens

$y$ is the number of positive multiples of $3$ that are less than or equal to $x$ .
The pair is $(a, 4)$ with $a$ a natural number.

Unknowns

The sum of all natural numbers $a$ with $y=4$ .

Constraints

The count only goes up when $x$ reaches a new multiple of $3$ , so several $x$ values can share the same $y$ .

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #2 Make a Systematic List

List the count as $x$ grows and watch the pattern: $y$ climbs by $1$ at each multiple of $3$ and stays put on the $2$ numbers after it. The values of $x$ with $y=4$ form a short stretch, which we list and add.

3 · Execute3 carry out the plan

1List the count near $y=4$

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

The positive multiples of

3

in order are

3, 6, 9, 12, 15, \dots

Up to

11

there are three of them;

12

brings the fourth;

15

brings the fifth.

x = 11 \to y = 3, \quad x = 12 \to y = 4, \quad x = 13 \to y = 4, \quad x = 14 \to y = 4, \quad x = 15 \to y = 5

Just count the multiples of

3

up to each

x

.

2Find the stretch where $y=4$

#5 Look for a Pattern 7.EE.B.4

Exactly four multiples of

3

,

3, 6, 9, 12

, are at most

a

when

a

has reached

12

but not yet

15

.

12 \le a < 15

Each count is held by a multiple of

3

and the

2

numbers right after it.

3Add the possible values

#2 Make a Systematic List 7.EE.B.4

The natural numbers in that stretch are

12

,

13

, and

14

.

12 + 13 + 14 = 39

All

3

give the same count, so all of them count.

Answer:

39

4 · Reviewdoes it hold up?

For $a=12$ the multiples of $3$ are $3, 6, 9, 12$ , which is four; for $a=14$ they are the same four; $a=11$ gives three and $a=15$ gives five. So exactly $12$ , $13$ , and $14$ work.

Another way: Write the count as the number of multiples of $3$ up to $a$ : it equals $4$ when $3 \times 4 \le a < 3 \times 5$ , giving $a$ from $12$ to $14$ directly.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the condition 'exactly four multiples of $3$' as the inequality $12 \le a < 15$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing multiples of $3$ up to a number.

💡Takeaway. When a rule counts things, many inputs can share an output -- find the whole stretch, not just one value.

Variant 5 medium

For two variables $x$ and $y$ , let $y$ be the number of positive even numbers less than or equal to $x$ . For the ordered pairs $(x, y)$ made this way, find the sum of all natural numbers $a$ for which $(a, 9)$ is one of the pairs.

Your answer

Show solution

1 · Understandwhat's really being asked

Each natural number $x$ is paired with how many positive even numbers are at most $x$ . Find every $x$ that gets paired with $9$ , and add them.

Givens

$y$ is the number of positive even numbers that are less than or equal to $x$ .
The pair is $(a, 9)$ with $a$ a natural number.

Unknowns

The sum of all natural numbers $a$ with $y=9$ .

Constraints

The count only goes up when $x$ reaches a new even number, so several $x$ values can share the same $y$ .

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #2 Make a Systematic List

List the count as $x$ grows and watch the pattern: $y$ climbs by $1$ at each even number and stays put on the odd number after it. The values of $x$ with $y=9$ form a short stretch, which we list and add.

3 · Execute3 carry out the plan

1List the count near $y=9$

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

The positive even numbers in order are

2, 4, 6, 8, 10, 12, 14, 16, 18, 20, \dots

Up to

17

there are eight of them;

18

brings the ninth;

20

brings the tenth.

x = 17 \to y = 8, \quad x = 18 \to y = 9, \quad x = 19 \to y = 9, \quad x = 20 \to y = 10

Just count the evens up to each

x

.

2Find the stretch where $y=9$

#5 Look for a Pattern 7.EE.B.4

Exactly nine evens,

2, 4, \dots, 18

, are at most

a

when

a

has reached

18

but not yet

20

.

18 \le a < 20

Each count is held by an even number and the odd number right after it.

3Add the possible values

#2 Make a Systematic List 7.EE.B.4

The natural numbers in that stretch are

18

and

19

.

18 + 19 = 37

Both give the same count, so both count.

Answer:

37

4 · Reviewdoes it hold up?

For $a=18$ the evens are $2, 4, \dots, 18$ , which is nine; for $a=19$ they are the same nine; $a=17$ gives eight and $a=20$ gives ten. So exactly $18$ and $19$ work.

Another way: Write the count as the number of multiples of $2$ up to $a$ : it equals $9$ when $2 \times 9 \le a < 2 \times 10$ , giving $a=18$ or $19$ directly.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the condition 'exactly nine evens' as the inequality $18 \le a < 20$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing multiples of $2$ up to a number.

💡Takeaway. When a rule counts things, many inputs can share an output -- find the whole stretch, not just one value.

Variant 6 medium

For two variables $x$ and $y$ , let $y$ be the number of positive multiples of $5$ less than or equal to $x$ . For the ordered pairs $(x, y)$ made this way, find the sum of all natural numbers $a$ for which $(a, 4)$ is one of the pairs.

Your answer

Show solution

1 · Understandwhat's really being asked

Each natural number $x$ is paired with how many positive multiples of $5$ are at most $x$ . Find every $x$ that gets paired with $4$ , and add them.

Givens

$y$ is the number of positive multiples of $5$ that are less than or equal to $x$ .
The pair is $(a, 4)$ with $a$ a natural number.

Unknowns

The sum of all natural numbers $a$ with $y=4$ .

Constraints

The count only goes up when $x$ reaches a new multiple of $5$ , so several $x$ values can share the same $y$ .

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #2 Make a Systematic List

List the count as $x$ grows and watch the pattern: $y$ climbs by $1$ at each multiple of $5$ and stays put on the $4$ numbers after it. The values of $x$ with $y=4$ form a short stretch, which we list and add.

3 · Execute3 carry out the plan

1List the count near $y=4$

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

The positive multiples of

5

in order are

5, 10, 15, 20, 25, \dots

Up to

19

there are three of them;

20

brings the fourth;

25

brings the fifth.

x = 19 \to y = 3, \quad x = 20 \to y = 4, \quad x = 21 \to y = 4, \quad x = 22 \to y = 4, \quad x = 23 \to y = 4, \quad x = 24 \to y = 4, \quad x = 25 \to y = 5

Just count the multiples of

5

up to each

x

.

2Find the stretch where $y=4$

#5 Look for a Pattern 7.EE.B.4

Exactly four multiples of

5

,

5, 10, 15, 20

, are at most

a

when

a

has reached

20

but not yet

25

.

20 \le a < 25

Each count is held by a multiple of

5

and the

4

numbers right after it.

3Add the possible values

#2 Make a Systematic List 7.EE.B.4

The natural numbers in that stretch are

20

,

21

,

22

,

23

, and

24

.

20 + 21 + 22 + 23 + 24 = 110

All

5

give the same count, so all of them count.

Answer:

110

4 · Reviewdoes it hold up?

For $a=20$ the multiples of $5$ are $5, 10, 15, 20$ , which is four; for $a=24$ they are the same four; $a=19$ gives three and $a=25$ gives five. So exactly $20$ , $21$ , $22$ , $23$ , and $24$ work.

Another way: Write the count as the number of multiples of $5$ up to $a$ : it equals $4$ when $5 \times 4 \le a < 5 \times 5$ , giving $a$ from $20$ to $24$ directly.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the condition 'exactly four multiples of $5$' as the inequality $20 \le a < 25$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing multiples of $5$ up to a number.

💡Takeaway. When a rule counts things, many inputs can share an output -- find the whole stretch, not just one value.

Variant 7 medium

For two variables $x$ and $y$ , let $y$ be the number of positive multiples of $4$ less than or equal to $x$ . For the ordered pairs $(x, y)$ made this way, find the sum of all natural numbers $a$ for which $(a, 5)$ is one of the pairs.

Your answer

Show solution

1 · Understandwhat's really being asked

Each natural number $x$ is paired with how many positive multiples of $4$ are at most $x$ . Find every $x$ that gets paired with $5$ , and add them.

Givens

$y$ is the number of positive multiples of $4$ that are less than or equal to $x$ .
The pair is $(a, 5)$ with $a$ a natural number.

Unknowns

The sum of all natural numbers $a$ with $y=5$ .

Constraints

The count only goes up when $x$ reaches a new multiple of $4$ , so several $x$ values can share the same $y$ .

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #2 Make a Systematic List

List the count as $x$ grows and watch the pattern: $y$ climbs by $1$ at each multiple of $4$ and stays put on the $3$ numbers after it. The values of $x$ with $y=5$ form a short stretch, which we list and add.

3 · Execute3 carry out the plan

1List the count near $y=5$

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

The positive multiples of

4

in order are

4, 8, 12, 16, 20, 24, \dots

Up to

19

there are four of them;

20

brings the fifth;

24

brings the sixth.

x = 19 \to y = 4, \quad x = 20 \to y = 5, \quad x = 21 \to y = 5, \quad x = 22 \to y = 5, \quad x = 23 \to y = 5, \quad x = 24 \to y = 6

Just count the multiples of

4

up to each

x

.

2Find the stretch where $y=5$

#5 Look for a Pattern 7.EE.B.4

Exactly five multiples of

4

,

4, 8, 12, 16, 20

, are at most

a

when

a

has reached

20

but not yet

24

.

20 \le a < 24

Each count is held by a multiple of

4

and the

3

numbers right after it.

3Add the possible values

#2 Make a Systematic List 7.EE.B.4

The natural numbers in that stretch are

20

,

21

,

22

, and

23

.

20 + 21 + 22 + 23 = 86

All

4

give the same count, so all of them count.

Answer:

86

4 · Reviewdoes it hold up?

For $a=20$ the multiples of $4$ are $4, 8, 12, 16, 20$ , which is five; for $a=23$ they are the same five; $a=19$ gives four and $a=24$ gives six. So exactly $20$ , $21$ , $22$ , and $23$ work.

Another way: Write the count as the number of multiples of $4$ up to $a$ : it equals $5$ when $4 \times 5 \le a < 4 \times 6$ , giving $a$ from $20$ to $23$ directly.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the condition 'exactly five multiples of $4$' as the inequality $20 \le a < 24$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing multiples of $4$ up to a number.

💡Takeaway. When a rule counts things, many inputs can share an output -- find the whole stretch, not just one value.

Variant 8 medium

For two variables $x$ and $y$ , let $y$ be the number of positive multiples of $3$ less than or equal to $x$ . For the ordered pairs $(x, y)$ made this way, find the sum of all natural numbers $a$ for which $(a, 6)$ is one of the pairs.

Your answer

Show solution

1 · Understandwhat's really being asked

Each natural number $x$ is paired with how many positive multiples of $3$ are at most $x$ . Find every $x$ that gets paired with $6$ , and add them.

Givens

$y$ is the number of positive multiples of $3$ that are less than or equal to $x$ .
The pair is $(a, 6)$ with $a$ a natural number.

Unknowns

The sum of all natural numbers $a$ with $y=6$ .

Constraints

The count only goes up when $x$ reaches a new multiple of $3$ , so several $x$ values can share the same $y$ .

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #2 Make a Systematic List

List the count as $x$ grows and watch the pattern: $y$ climbs by $1$ at each multiple of $3$ and stays put on the $2$ numbers after it. The values of $x$ with $y=6$ form a short stretch, which we list and add.

3 · Execute3 carry out the plan

1List the count near $y=6$

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

The positive multiples of

3

in order are

3, 6, 9, 12, 15, 18, 21, \dots

Up to

17

there are five of them;

18

brings the sixth;

21

brings the seventh.

x = 17 \to y = 5, \quad x = 18 \to y = 6, \quad x = 19 \to y = 6, \quad x = 20 \to y = 6, \quad x = 21 \to y = 7

Just count the multiples of

3

up to each

x

.

2Find the stretch where $y=6$

#5 Look for a Pattern 7.EE.B.4

Exactly six multiples of

3

,

3, 6, 9, 12, 15, 18

, are at most

a

when

a

has reached

18

but not yet

21

.

18 \le a < 21

Each count is held by a multiple of

3

and the

2

numbers right after it.

3Add the possible values

#2 Make a Systematic List 7.EE.B.4

The natural numbers in that stretch are

18

,

19

, and

20

.

18 + 19 + 20 = 57

All

3

give the same count, so all of them count.

Answer:

57

4 · Reviewdoes it hold up?

For $a=18$ the multiples of $3$ are $3, 6, 9, 12, 15, 18$ , which is six; for $a=20$ they are the same six; $a=17$ gives five and $a=21$ gives seven. So exactly $18$ , $19$ , and $20$ work.

Another way: Write the count as the number of multiples of $3$ up to $a$ : it equals $6$ when $3 \times 6 \le a < 3 \times 7$ , giving $a$ from $18$ to $20$ directly.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the condition 'exactly six multiples of $3$' as the inequality $18 \le a < 21$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing multiples of $3$ up to a number.

💡Takeaway. When a rule counts things, many inputs can share an output -- find the whole stretch, not just one value.

Variant 9 hard

For two variables $x$ and $y$ , let $y$ be the number of positive multiples of $5$ less than or equal to $x$ . For the ordered pairs $(x, y)$ made this way, find the sum of all natural numbers $a$ for which $(a, 6)$ is one of the pairs.

Your answer

Show solution

1 · Understandwhat's really being asked

Each natural number $x$ is paired with how many positive multiples of $5$ are at most $x$ . Find every $x$ that gets paired with $6$ , and add them.

Givens

$y$ is the number of positive multiples of $5$ that are less than or equal to $x$ .
The pair is $(a, 6)$ with $a$ a natural number.

Unknowns

The sum of all natural numbers $a$ with $y=6$ .

Constraints

The count only goes up when $x$ reaches a new multiple of $5$ , so several $x$ values can share the same $y$ .

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #2 Make a Systematic List

List the count as $x$ grows and watch the pattern: $y$ climbs by $1$ at each multiple of $5$ and stays put on the $4$ numbers after it. The values of $x$ with $y=6$ form a short stretch, which we list and add.

3 · Execute3 carry out the plan

1List the count near $y=6$

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

The positive multiples of

5

in order are

5, 10, 15, 20, 25, 30, 35, \dots

Up to

29

there are five of them;

30

brings the sixth;

35

brings the seventh.

x = 29 \to y = 5, \quad x = 30 \to y = 6, \quad x = 31 \to y = 6, \quad x = 32 \to y = 6, \quad x = 33 \to y = 6, \quad x = 34 \to y = 6, \quad x = 35 \to y = 7

Just count the multiples of

5

up to each

x

.

2Find the stretch where $y=6$

#5 Look for a Pattern 7.EE.B.4

Exactly six multiples of

5

,

5, 10, 15, 20, 25, 30

, are at most

a

when

a

has reached

30

but not yet

35

.

30 \le a < 35

Each count is held by a multiple of

5

and the

4

numbers right after it.

3Add the possible values

#2 Make a Systematic List 7.EE.B.4

The natural numbers in that stretch are

30

,

31

,

32

,

33

, and

34

.

30 + 31 + 32 + 33 + 34 = 160

All

5

give the same count, so all of them count.

Answer:

160

4 · Reviewdoes it hold up?

For $a=30$ the multiples of $5$ are $5, 10, 15, 20, 25, 30$ , which is six; for $a=34$ they are the same six; $a=29$ gives five and $a=35$ gives seven. So exactly $30$ , $31$ , $32$ , $33$ , and $34$ work.

Another way: Write the count as the number of multiples of $5$ up to $a$ : it equals $6$ when $5 \times 6 \le a < 5 \times 7$ , giving $a$ from $30$ to $34$ directly.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the condition 'exactly six multiples of $5$' as the inequality $30 \le a < 35$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing multiples of $5$ up to a number.

💡Takeaway. When a rule counts things, many inputs can share an output -- find the whole stretch, not just one value.

Variant 10 hard

For two variables $x$ and $y$ , let $y$ be the number of positive even numbers less than or equal to $x$ . For the ordered pairs $(x, y)$ made this way, find the sum of all natural numbers $a$ for which $(a, 12)$ is one of the pairs.

Your answer

Show solution

1 · Understandwhat's really being asked

Each natural number $x$ is paired with how many positive even numbers are at most $x$ . Find every $x$ that gets paired with $12$ , and add them.

Givens

$y$ is the number of positive even numbers that are less than or equal to $x$ .
The pair is $(a, 12)$ with $a$ a natural number.

Unknowns

The sum of all natural numbers $a$ with $y=12$ .

Constraints

The count only goes up when $x$ reaches a new even number, so several $x$ values can share the same $y$ .

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #2 Make a Systematic List

List the count as $x$ grows and watch the pattern: $y$ climbs by $1$ at each even number and stays put on the odd number after it. The values of $x$ with $y=12$ form a short stretch, which we list and add.

3 · Execute3 carry out the plan

1List the count near $y=12$

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

The positive even numbers in order are

2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, \dots

Up to

23

there are eleven of them;

24

brings the twelfth;

26

brings the thirteenth.

x = 23 \to y = 11, \quad x = 24 \to y = 12, \quad x = 25 \to y = 12, \quad x = 26 \to y = 13

Just count the evens up to each

x

.

2Find the stretch where $y=12$

#5 Look for a Pattern 7.EE.B.4

Exactly twelve evens,

2, 4, \dots, 24

, are at most

a

when

a

has reached

24

but not yet

26

.

24 \le a < 26

Each count is held by an even number and the odd number right after it.

3Add the possible values

#2 Make a Systematic List 7.EE.B.4

The natural numbers in that stretch are

24

and

25

.

24 + 25 = 49

Both give the same count, so both count.

Answer:

49

4 · Reviewdoes it hold up?

For $a=24$ the evens are $2, 4, \dots, 24$ , which is twelve; for $a=25$ they are the same twelve; $a=23$ gives eleven and $a=26$ gives thirteen. So exactly $24$ and $25$ work.

Another way: Write the count as the number of multiples of $2$ up to $a$ : it equals $12$ when $2 \times 12 \le a < 2 \times 13$ , giving $a=24$ or $25$ directly.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the condition 'exactly twelve evens' as the inequality $24 \le a < 26$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing multiples of $2$ up to a number.

💡Takeaway. When a rule counts things, many inputs can share an output -- find the whole stretch, not just one value.

Variant 11 hard

For two variables $x$ and $y$ , let $y$ be the number of positive multiples of $4$ less than or equal to $x$ . For the ordered pairs $(x, y)$ made this way, find the sum of all natural numbers $a$ for which $(a, 8)$ is one of the pairs.

Your answer

Show solution

1 · Understandwhat's really being asked

Each natural number $x$ is paired with how many positive multiples of $4$ are at most $x$ . Find every $x$ that gets paired with $8$ , and add them.

Givens

$y$ is the number of positive multiples of $4$ that are less than or equal to $x$ .
The pair is $(a, 8)$ with $a$ a natural number.

Unknowns

The sum of all natural numbers $a$ with $y=8$ .

Constraints

The count only goes up when $x$ reaches a new multiple of $4$ , so several $x$ values can share the same $y$ .

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #2 Make a Systematic List

List the count as $x$ grows and watch the pattern: $y$ climbs by $1$ at each multiple of $4$ and stays put on the $3$ numbers after it. The values of $x$ with $y=8$ form a short stretch, which we list and add.

3 · Execute3 carry out the plan

1List the count near $y=8$

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

The positive multiples of

4

in order are

4, 8, 12, 16, 20, 24, 28, 32, 36, \dots

Up to

31

there are seven of them;

32

brings the eighth;

36

brings the ninth.

x = 31 \to y = 7, \quad x = 32 \to y = 8, \quad x = 33 \to y = 8, \quad x = 34 \to y = 8, \quad x = 35 \to y = 8, \quad x = 36 \to y = 9

Just count the multiples of

4

up to each

x

.

2Find the stretch where $y=8$

#5 Look for a Pattern 7.EE.B.4

Exactly eight multiples of

4

,

4, 8, \dots, 32

, are at most

a

when

a

has reached

32

but not yet

36

.

32 \le a < 36

Each count is held by a multiple of

4

and the

3

numbers right after it.

3Add the possible values

#2 Make a Systematic List 7.EE.B.4

The natural numbers in that stretch are

32

,

33

,

34

, and

35

.

32 + 33 + 34 + 35 = 134

All

4

give the same count, so all of them count.

Answer:

134

4 · Reviewdoes it hold up?

For $a=32$ the multiples of $4$ are $4, 8, \dots, 32$ , which is eight; for $a=35$ they are the same eight; $a=31$ gives seven and $a=36$ gives nine. So exactly $32$ , $33$ , $34$ , and $35$ work.

Another way: Write the count as the number of multiples of $4$ up to $a$ : it equals $8$ when $4 \times 8 \le a < 4 \times 9$ , giving $a$ from $32$ to $35$ directly.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the condition 'exactly eight multiples of $4$' as the inequality $32 \le a < 36$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing multiples of $4$ up to a number.

💡Takeaway. When a rule counts things, many inputs can share an output -- find the whole stretch, not just one value.

Variant 12 hard

For two variables $x$ and $y$ , let $y$ be the number of positive multiples of $3$ less than or equal to $x$ . For the ordered pairs $(x, y)$ made this way, find the sum of all natural numbers $a$ for which $(a, 10)$ is one of the pairs.

Your answer

Show solution

1 · Understandwhat's really being asked

Each natural number $x$ is paired with how many positive multiples of $3$ are at most $x$ . Find every $x$ that gets paired with $10$ , and add them.

Givens

$y$ is the number of positive multiples of $3$ that are less than or equal to $x$ .
The pair is $(a, 10)$ with $a$ a natural number.

Unknowns

The sum of all natural numbers $a$ with $y=10$ .

Constraints

The count only goes up when $x$ reaches a new multiple of $3$ , so several $x$ values can share the same $y$ .

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #2 Make a Systematic List

List the count as $x$ grows and watch the pattern: $y$ climbs by $1$ at each multiple of $3$ and stays put on the $2$ numbers after it. The values of $x$ with $y=10$ form a short stretch, which we list and add.

3 · Execute3 carry out the plan

1List the count near $y=10$

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

The positive multiples of

3

in order are

3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, \dots

Up to

29

there are nine of them;

30

brings the tenth;

33

brings the eleventh.

x = 29 \to y = 9, \quad x = 30 \to y = 10, \quad x = 31 \to y = 10, \quad x = 32 \to y = 10, \quad x = 33 \to y = 11

Just count the multiples of

3

up to each

x

.

2Find the stretch where $y=10$

#5 Look for a Pattern 7.EE.B.4

Exactly ten multiples of

3

,

3, 6, \dots, 30

, are at most

a

when

a

has reached

30

but not yet

33

.

30 \le a < 33

Each count is held by a multiple of

3

and the

2

numbers right after it.

3Add the possible values

#2 Make a Systematic List 7.EE.B.4

The natural numbers in that stretch are

30

,

31

, and

32

.

30 + 31 + 32 = 93

All

3

give the same count, so all of them count.

Answer:

93

4 · Reviewdoes it hold up?

For $a=30$ the multiples of $3$ are $3, 6, \dots, 30$ , which is ten; for $a=32$ they are the same ten; $a=29$ gives nine and $a=33$ gives eleven. So exactly $30$ , $31$ , and $32$ work.

Another way: Write the count as the number of multiples of $3$ up to $a$ : it equals $10$ when $3 \times 10 \le a < 3 \times 11$ , giving $a$ from $30$ to $32$ directly.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the condition 'exactly ten multiples of $3$' as the inequality $30 \le a < 33$.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing multiples of $3$ up to a number.

💡Takeaway. When a rule counts things, many inputs can share an output -- find the whole stretch, not just one value.

Run a counting rule backward to every input · 12 practice problems

Generated variants — 12

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