Practice · Same shortfall turns remainders into one LCM · Sensim Math

Variant 1 easy

A natural number leaves a remainder of $3$ when divided by $4$ , and a remainder of $5$ when divided by $6$ . What is the remainder when this number is divided by $12$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A number leaves $3$ when divided by $4$ and $5$ when divided by $6$ . Find what it leaves when divided by $12$ .

Givens

Divided by $4$ , the remainder is $3$ .
Divided by $6$ , the remainder is $5$ .

Unknowns

The remainder when the number is divided by $12$ .

Constraints

The two remainders are different, so the number is not simply a common multiple plus one fixed remainder.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#2 Make a Systematic List

Instead of looking at what is left over, look at what is missing. Both remainders are exactly $1$ short of the divisor, so adding $1$ makes the number a common multiple of $4$ and $6$ .

3 · Execute4 carry out the plan

1Look at the shortfall

#16 Count the Complement 7.EE.A.2

A remainder of

3

out of

4

is

1

short of the next multiple of

4

; a remainder of

5

out of

6

is

1

short of the next multiple of

6

.

x = 4a + 3 \Rightarrow x + 1 = 4(a + 1), \quad x = 6b + 5 \Rightarrow x + 1 = 6(b + 1)

Adding

1

makes

x

a multiple of both

4

and

6

.

2Use the least common multiple

#13 Convert to Algebra 6.NS.B.4

A common multiple of

4

and

6

is a multiple of their least common multiple,

12

.

x + 1 = 12k

x

is

1

less than a multiple of

12

.

3Read off the remainder

#13 Convert to Algebra 7.EE.A.2

Rewrite

x

so that the part left over is between

0

and

11

.

x = 12k - 1 = 12(k - 1) + 11

Divided by

12

, the remainder is

11

.

4List a few to confirm

#2 Make a Systematic List 6.NS.B.4

The numbers are

11, 23, 35, \ldots

: each is

11

more than a multiple of

12

.

11 = 12 \times 0 + 11, \quad 23 = 12 \times 1 + 11

Every such number leaves

11

.

Answer: 11

4 · Reviewdoes it hold up?

Take $23$ : $23 = 4 \times 5 + 3$ and $23 = 6 \times 3 + 5$ , as required, and $23 = 12 \times 1 + 11$ .

Another way: List numbers that leave $5$ when divided by $6$ ( $5, 11, 17, \ldots$ ) and keep the first that leaves $3$ when divided by $4$ : that is $11$ , and adding $12$ keeps both remainders, so the remainder by $12$ is always $11$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the LCM of $4$ and $6$, which is $12$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting $x = 4a + 3$ as $x + 1 = 4(a + 1)$ and $12k - 1$ as $12(k - 1) + 11$.

💡Takeaway. When the remainders differ, check the shortfall -- if it is the same, add it back and use the LCM.

Variant 2 easy

A natural number leaves a remainder of $2$ when divided by $3$ , and a remainder of $4$ when divided by $5$ . What is the remainder when this number is divided by $15$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A number leaves $2$ when divided by $3$ and $4$ when divided by $5$ . Find what it leaves when divided by $15$ .

Givens

Divided by $3$ , the remainder is $2$ .
Divided by $5$ , the remainder is $4$ .

Unknowns

The remainder when the number is divided by $15$ .

Constraints

The two remainders are different, so the number is not simply a common multiple plus one fixed remainder.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#2 Make a Systematic List

Instead of looking at what is left over, look at what is missing. Both remainders are exactly $1$ short of the divisor, so adding $1$ makes the number a common multiple of $3$ and $5$ .

3 · Execute4 carry out the plan

1Look at the shortfall

#16 Count the Complement 7.EE.A.2

A remainder of

2

out of

3

is

1

short of the next multiple of

3

; a remainder of

4

out of

5

is

1

short of the next multiple of

5

.

x = 3a + 2 \Rightarrow x + 1 = 3(a + 1), \quad x = 5b + 4 \Rightarrow x + 1 = 5(b + 1)

Adding

1

makes

x

a multiple of both

3

and

5

.

2Use the least common multiple

#13 Convert to Algebra 6.NS.B.4

A common multiple of

3

and

5

is a multiple of their least common multiple,

15

.

x + 1 = 15k

x

is

1

less than a multiple of

15

.

3Read off the remainder

#13 Convert to Algebra 7.EE.A.2

Rewrite

x

so that the part left over is between

0

and

14

.

x = 15k - 1 = 15(k - 1) + 14

Divided by

15

, the remainder is

14

.

4List a few to confirm

#2 Make a Systematic List 6.NS.B.4

The numbers are

14, 29, 44, \ldots

: each is

14

more than a multiple of

15

.

14 = 15 \times 0 + 14, \quad 29 = 15 \times 1 + 14

Every such number leaves

14

.

Answer: 14

4 · Reviewdoes it hold up?

Take $29$ : $29 = 3 \times 9 + 2$ and $29 = 5 \times 5 + 4$ , as required, and $29 = 15 \times 1 + 14$ .

Another way: List numbers that leave $4$ when divided by $5$ ( $4, 9, 14, \ldots$ ) and keep the first that leaves $2$ when divided by $3$ : that is $14$ , and adding $15$ keeps both remainders, so the remainder by $15$ is always $14$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the LCM of $3$ and $5$, which is $15$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting $x = 3a + 2$ as $x + 1 = 3(a + 1)$ and $15k - 1$ as $15(k - 1) + 14$.

💡Takeaway. When the remainders differ, check the shortfall -- if it is the same, add it back and use the LCM.

Variant 3 easy

A natural number leaves a remainder of $3$ when divided by $6$ , and a remainder of $6$ when divided by $9$ . What is the remainder when this number is divided by $18$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A number leaves $3$ when divided by $6$ and $6$ when divided by $9$ . Find what it leaves when divided by $18$ .

Givens

Divided by $6$ , the remainder is $3$ .
Divided by $9$ , the remainder is $6$ .

Unknowns

The remainder when the number is divided by $18$ .

Constraints

The two remainders are different, so the number is not simply a common multiple plus one fixed remainder.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#2 Make a Systematic List

Instead of looking at what is left over, look at what is missing. Both remainders are exactly $3$ short of the divisor, so adding $3$ makes the number a common multiple of $6$ and $9$ .

3 · Execute4 carry out the plan

1Look at the shortfall

#16 Count the Complement 7.EE.A.2

A remainder of

3

out of

6

is

3

short of the next multiple of

6

; a remainder of

6

out of

9

is

3

short of the next multiple of

9

.

x = 6a + 3 \Rightarrow x + 3 = 6(a + 1), \quad x = 9b + 6 \Rightarrow x + 3 = 9(b + 1)

Adding

3

makes

x

a multiple of both

6

and

9

.

2Use the least common multiple

#13 Convert to Algebra 6.NS.B.4

A common multiple of

6

and

9

is a multiple of their least common multiple,

18

.

x + 3 = 18k

x

is

3

less than a multiple of

18

.

3Read off the remainder

#13 Convert to Algebra 7.EE.A.2

Rewrite

x

so that the part left over is between

0

and

17

.

x = 18k - 3 = 18(k - 1) + 15

Divided by

18

, the remainder is

15

.

4List a few to confirm

#2 Make a Systematic List 6.NS.B.4

The numbers are

15, 33, 51, \ldots

: each is

15

more than a multiple of

18

.

15 = 18 \times 0 + 15, \quad 33 = 18 \times 1 + 15

Every such number leaves

15

.

Answer: 15

4 · Reviewdoes it hold up?

Take $33$ : $33 = 6 \times 5 + 3$ and $33 = 9 \times 3 + 6$ , as required, and $33 = 18 \times 1 + 15$ .

Another way: List numbers that leave $6$ when divided by $9$ ( $6, 15, 24, \ldots$ ) and keep the first that leaves $3$ when divided by $6$ : that is $15$ , and adding $18$ keeps both remainders, so the remainder by $18$ is always $15$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the LCM of $6$ and $9$, which is $18$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting $x = 6a + 3$ as $x + 3 = 6(a + 1)$ and $18k - 3$ as $18(k - 1) + 15$.

💡Takeaway. When the remainders differ, check the shortfall -- if it is the same, add it back and use the LCM.

Variant 4 easy

A natural number leaves a remainder of $1$ when divided by $3$ , and a remainder of $5$ when divided by $7$ . What is the remainder when this number is divided by $21$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A number leaves $1$ when divided by $3$ and $5$ when divided by $7$ . Find what it leaves when divided by $21$ .

Givens

Divided by $3$ , the remainder is $1$ .
Divided by $7$ , the remainder is $5$ .

Unknowns

The remainder when the number is divided by $21$ .

Constraints

The two remainders are different, so the number is not simply a common multiple plus one fixed remainder.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#2 Make a Systematic List

Instead of looking at what is left over, look at what is missing. Both remainders are exactly $2$ short of the divisor, so adding $2$ makes the number a common multiple of $3$ and $7$ .

3 · Execute4 carry out the plan

1Look at the shortfall

#16 Count the Complement 7.EE.A.2

A remainder of

1

out of

3

is

2

short of the next multiple of

3

; a remainder of

5

out of

7

is

2

short of the next multiple of

7

.

x = 3a + 1 \Rightarrow x + 2 = 3(a + 1), \quad x = 7b + 5 \Rightarrow x + 2 = 7(b + 1)

Adding

2

makes

x

a multiple of both

3

and

7

.

2Use the least common multiple

#13 Convert to Algebra 6.NS.B.4

A common multiple of

3

and

7

is a multiple of their least common multiple,

21

.

x + 2 = 21k

x

is

2

less than a multiple of

21

.

3Read off the remainder

#13 Convert to Algebra 7.EE.A.2

Rewrite

x

so that the part left over is between

0

and

20

.

x = 21k - 2 = 21(k - 1) + 19

Divided by

21

, the remainder is

19

.

4List a few to confirm

#2 Make a Systematic List 6.NS.B.4

The numbers are

19, 40, 61, \ldots

: each is

19

more than a multiple of

21

.

19 = 21 \times 0 + 19, \quad 40 = 21 \times 1 + 19

Every such number leaves

19

.

Answer: 19

4 · Reviewdoes it hold up?

Take $40$ : $40 = 3 \times 13 + 1$ and $40 = 7 \times 5 + 5$ , as required, and $40 = 21 \times 1 + 19$ .

Another way: List numbers that leave $5$ when divided by $7$ ( $5, 12, 19, \ldots$ ) and keep the first that leaves $1$ when divided by $3$ : that is $19$ , and adding $21$ keeps both remainders, so the remainder by $21$ is always $19$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the LCM of $3$ and $7$, which is $21$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting $x = 3a + 1$ as $x + 2 = 3(a + 1)$ and $21k - 2$ as $21(k - 1) + 19$.

💡Takeaway. When the remainders differ, check the shortfall -- if it is the same, add it back and use the LCM.

Variant 5 medium

A natural number leaves a remainder of $4$ when divided by $6$ , and a remainder of $6$ when divided by $8$ . What is the remainder when this number is divided by $24$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A number leaves $4$ when divided by $6$ and $6$ when divided by $8$ . Find what it leaves when divided by $24$ .

Givens

Divided by $6$ , the remainder is $4$ .
Divided by $8$ , the remainder is $6$ .

Unknowns

The remainder when the number is divided by $24$ .

Constraints

The two remainders are different, so the number is not simply a common multiple plus one fixed remainder.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#2 Make a Systematic List

Instead of looking at what is left over, look at what is missing. Both remainders are exactly $2$ short of the divisor, so adding $2$ makes the number a common multiple of $6$ and $8$ .

3 · Execute4 carry out the plan

1Look at the shortfall

#16 Count the Complement 7.EE.A.2

A remainder of

4

out of

6

is

2

short of the next multiple of

6

; a remainder of

6

out of

8

is

2

short of the next multiple of

8

.

x = 6a + 4 \Rightarrow x + 2 = 6(a + 1), \quad x = 8b + 6 \Rightarrow x + 2 = 8(b + 1)

Adding

2

makes

x

a multiple of both

6

and

8

.

2Use the least common multiple

#13 Convert to Algebra 6.NS.B.4

A common multiple of

6

and

8

is a multiple of their least common multiple,

24

.

x + 2 = 24k

x

is

2

less than a multiple of

24

.

3Read off the remainder

#13 Convert to Algebra 7.EE.A.2

Rewrite

x

so that the part left over is between

0

and

23

.

x = 24k - 2 = 24(k - 1) + 22

Divided by

24

, the remainder is

22

.

4List a few to confirm

#2 Make a Systematic List 6.NS.B.4

The numbers are

22, 46, 70, \ldots

: each is

22

more than a multiple of

24

.

22 = 24 \times 0 + 22, \quad 46 = 24 \times 1 + 22

Every such number leaves

22

.

Answer: 22

4 · Reviewdoes it hold up?

Take $46$ : $46 = 6 \times 7 + 4$ and $46 = 8 \times 5 + 6$ , as required, and $46 = 24 \times 1 + 22$ .

Another way: List numbers that leave $6$ when divided by $8$ ( $6, 14, 22, \ldots$ ) and keep the first that leaves $4$ when divided by $6$ : that is $22$ , and adding $24$ keeps both remainders, so the remainder by $24$ is always $22$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the LCM of $6$ and $8$, which is $24$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting $x = 6a + 4$ as $x + 2 = 6(a + 1)$ and $24k - 2$ as $24(k - 1) + 22$.

💡Takeaway. When the remainders differ, check the shortfall -- if it is the same, add it back and use the LCM.

Variant 6 medium

A natural number leaves a remainder of $3$ when divided by $4$ , and a remainder of $6$ when divided by $7$ . What is the remainder when this number is divided by $28$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A number leaves $3$ when divided by $4$ and $6$ when divided by $7$ . Find what it leaves when divided by $28$ .

Givens

Divided by $4$ , the remainder is $3$ .
Divided by $7$ , the remainder is $6$ .

Unknowns

The remainder when the number is divided by $28$ .

Constraints

The two remainders are different, so the number is not simply a common multiple plus one fixed remainder.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#2 Make a Systematic List

Instead of looking at what is left over, look at what is missing. Both remainders are exactly $1$ short of the divisor, so adding $1$ makes the number a common multiple of $4$ and $7$ .

3 · Execute4 carry out the plan

1Look at the shortfall

#16 Count the Complement 7.EE.A.2

A remainder of

3

out of

4

is

1

short of the next multiple of

4

; a remainder of

6

out of

7

is

1

short of the next multiple of

7

.

x = 4a + 3 \Rightarrow x + 1 = 4(a + 1), \quad x = 7b + 6 \Rightarrow x + 1 = 7(b + 1)

Adding

1

makes

x

a multiple of both

4

and

7

.

2Use the least common multiple

#13 Convert to Algebra 6.NS.B.4

A common multiple of

4

and

7

is a multiple of their least common multiple,

28

.

x + 1 = 28k

x

is

1

less than a multiple of

28

.

3Read off the remainder

#13 Convert to Algebra 7.EE.A.2

Rewrite

x

so that the part left over is between

0

and

27

.

x = 28k - 1 = 28(k - 1) + 27

Divided by

28

, the remainder is

27

.

4List a few to confirm

#2 Make a Systematic List 6.NS.B.4

The numbers are

27, 55, 83, \ldots

: each is

27

more than a multiple of

28

.

27 = 28 \times 0 + 27, \quad 55 = 28 \times 1 + 27

Every such number leaves

27

.

Answer: 27

4 · Reviewdoes it hold up?

Take $55$ : $55 = 4 \times 13 + 3$ and $55 = 7 \times 7 + 6$ , as required, and $55 = 28 \times 1 + 27$ .

Another way: List numbers that leave $6$ when divided by $7$ ( $6, 13, 20, \ldots$ ) and keep the first that leaves $3$ when divided by $4$ : that is $27$ , and adding $28$ keeps both remainders, so the remainder by $28$ is always $27$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the LCM of $4$ and $7$, which is $28$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting $x = 4a + 3$ as $x + 1 = 4(a + 1)$ and $28k - 1$ as $28(k - 1) + 27$.

💡Takeaway. When the remainders differ, check the shortfall -- if it is the same, add it back and use the LCM.

Variant 7 medium

A natural number leaves a remainder of $4$ when divided by $5$ , and a remainder of $5$ when divided by $6$ . What is the remainder when this number is divided by $30$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A number leaves $4$ when divided by $5$ and $5$ when divided by $6$ . Find what it leaves when divided by $30$ .

Givens

Divided by $5$ , the remainder is $4$ .
Divided by $6$ , the remainder is $5$ .

Unknowns

The remainder when the number is divided by $30$ .

Constraints

The two remainders are different, so the number is not simply a common multiple plus one fixed remainder.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#2 Make a Systematic List

Instead of looking at what is left over, look at what is missing. Both remainders are exactly $1$ short of the divisor, so adding $1$ makes the number a common multiple of $5$ and $6$ .

3 · Execute4 carry out the plan

1Look at the shortfall

#16 Count the Complement 7.EE.A.2

A remainder of

4

out of

5

is

1

short of the next multiple of

5

; a remainder of

5

out of

6

is

1

short of the next multiple of

6

.

x = 5a + 4 \Rightarrow x + 1 = 5(a + 1), \quad x = 6b + 5 \Rightarrow x + 1 = 6(b + 1)

Adding

1

makes

x

a multiple of both

5

and

6

.

2Use the least common multiple

#13 Convert to Algebra 6.NS.B.4

A common multiple of

5

and

6

is a multiple of their least common multiple,

30

.

x + 1 = 30k

x

is

1

less than a multiple of

30

.

3Read off the remainder

#13 Convert to Algebra 7.EE.A.2

Rewrite

x

so that the part left over is between

0

and

29

.

x = 30k - 1 = 30(k - 1) + 29

Divided by

30

, the remainder is

29

.

4List a few to confirm

#2 Make a Systematic List 6.NS.B.4

The numbers are

29, 59, 89, \ldots

: each is

29

more than a multiple of

30

.

29 = 30 \times 0 + 29, \quad 59 = 30 \times 1 + 29

Every such number leaves

29

.

Answer: 29

4 · Reviewdoes it hold up?

Take $59$ : $59 = 5 \times 11 + 4$ and $59 = 6 \times 9 + 5$ , as required, and $59 = 30 \times 1 + 29$ .

Another way: List numbers that leave $5$ when divided by $6$ ( $5, 11, 17, \ldots$ ) and keep the first that leaves $4$ when divided by $5$ : that is $29$ , and adding $30$ keeps both remainders, so the remainder by $30$ is always $29$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the LCM of $5$ and $6$, which is $30$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting $x = 5a + 4$ as $x + 1 = 5(a + 1)$ and $30k - 1$ as $30(k - 1) + 29$.

💡Takeaway. When the remainders differ, check the shortfall -- if it is the same, add it back and use the LCM.

Variant 8 medium

A natural number leaves a remainder of $3$ when divided by $5$ , and a remainder of $5$ when divided by $7$ . What is the remainder when this number is divided by $35$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A number leaves $3$ when divided by $5$ and $5$ when divided by $7$ . Find what it leaves when divided by $35$ .

Givens

Divided by $5$ , the remainder is $3$ .
Divided by $7$ , the remainder is $5$ .

Unknowns

The remainder when the number is divided by $35$ .

Constraints

The two remainders are different, so the number is not simply a common multiple plus one fixed remainder.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#2 Make a Systematic List

Instead of looking at what is left over, look at what is missing. Both remainders are exactly $2$ short of the divisor, so adding $2$ makes the number a common multiple of $5$ and $7$ .

3 · Execute4 carry out the plan

1Look at the shortfall

#16 Count the Complement 7.EE.A.2

A remainder of

3

out of

5

is

2

short of the next multiple of

5

; a remainder of

5

out of

7

is

2

short of the next multiple of

7

.

x = 5a + 3 \Rightarrow x + 2 = 5(a + 1), \quad x = 7b + 5 \Rightarrow x + 2 = 7(b + 1)

Adding

2

makes

x

a multiple of both

5

and

7

.

2Use the least common multiple

#13 Convert to Algebra 6.NS.B.4

A common multiple of

5

and

7

is a multiple of their least common multiple,

35

.

x + 2 = 35k

x

is

2

less than a multiple of

35

.

3Read off the remainder

#13 Convert to Algebra 7.EE.A.2

Rewrite

x

so that the part left over is between

0

and

34

.

x = 35k - 2 = 35(k - 1) + 33

Divided by

35

, the remainder is

33

.

4List a few to confirm

#2 Make a Systematic List 6.NS.B.4

The numbers are

33, 68, 103, \ldots

: each is

33

more than a multiple of

35

.

33 = 35 \times 0 + 33, \quad 68 = 35 \times 1 + 33

Every such number leaves

33

.

Answer: 33

4 · Reviewdoes it hold up?

Take $68$ : $68 = 5 \times 13 + 3$ and $68 = 7 \times 9 + 5$ , as required, and $68 = 35 \times 1 + 33$ .

Another way: List numbers that leave $5$ when divided by $7$ ( $5, 12, 19, \ldots$ ) and keep the first that leaves $3$ when divided by $5$ : that is $33$ , and adding $35$ keeps both remainders, so the remainder by $35$ is always $33$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the LCM of $5$ and $7$, which is $35$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting $x = 5a + 3$ as $x + 2 = 5(a + 1)$ and $35k - 2$ as $35(k - 1) + 33$.

💡Takeaway. When the remainders differ, check the shortfall -- if it is the same, add it back and use the LCM.

Variant 9 hard

A natural number leaves a remainder of $3$ when divided by $4$ , and a remainder of $8$ when divided by $9$ . What is the remainder when this number is divided by $36$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A number leaves $3$ when divided by $4$ and $8$ when divided by $9$ . Find what it leaves when divided by $36$ .

Givens

Divided by $4$ , the remainder is $3$ .
Divided by $9$ , the remainder is $8$ .

Unknowns

The remainder when the number is divided by $36$ .

Constraints

The two remainders are different, so the number is not simply a common multiple plus one fixed remainder.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#2 Make a Systematic List

Instead of looking at what is left over, look at what is missing. Both remainders are exactly $1$ short of the divisor, so adding $1$ makes the number a common multiple of $4$ and $9$ .

3 · Execute4 carry out the plan

1Look at the shortfall

#16 Count the Complement 7.EE.A.2

A remainder of

3

out of

4

is

1

short of the next multiple of

4

; a remainder of

8

out of

9

is

1

short of the next multiple of

9

.

x = 4a + 3 \Rightarrow x + 1 = 4(a + 1), \quad x = 9b + 8 \Rightarrow x + 1 = 9(b + 1)

Adding

1

makes

x

a multiple of both

4

and

9

.

2Use the least common multiple

#13 Convert to Algebra 6.NS.B.4

A common multiple of

4

and

9

is a multiple of their least common multiple,

36

.

x + 1 = 36k

x

is

1

less than a multiple of

36

.

3Read off the remainder

#13 Convert to Algebra 7.EE.A.2

Rewrite

x

so that the part left over is between

0

and

35

.

x = 36k - 1 = 36(k - 1) + 35

Divided by

36

, the remainder is

35

.

4List a few to confirm

#2 Make a Systematic List 6.NS.B.4

The numbers are

35, 71, 107, \ldots

: each is

35

more than a multiple of

36

.

35 = 36 \times 0 + 35, \quad 71 = 36 \times 1 + 35

Every such number leaves

35

.

Answer: 35

4 · Reviewdoes it hold up?

Take $71$ : $71 = 4 \times 17 + 3$ and $71 = 9 \times 7 + 8$ , as required, and $71 = 36 \times 1 + 35$ .

Another way: List numbers that leave $8$ when divided by $9$ ( $8, 17, 26, \ldots$ ) and keep the first that leaves $3$ when divided by $4$ : that is $35$ , and adding $36$ keeps both remainders, so the remainder by $36$ is always $35$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the LCM of $4$ and $9$, which is $36$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting $x = 4a + 3$ as $x + 1 = 4(a + 1)$ and $36k - 1$ as $36(k - 1) + 35$.

💡Takeaway. When the remainders differ, check the shortfall -- if it is the same, add it back and use the LCM.

Variant 10 hard

A natural number leaves a remainder of $7$ when divided by $9$ , and a remainder of $10$ when divided by $12$ . What is the remainder when this number is divided by $36$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A number leaves $7$ when divided by $9$ and $10$ when divided by $12$ . Find what it leaves when divided by $36$ .

Givens

Divided by $9$ , the remainder is $7$ .
Divided by $12$ , the remainder is $10$ .

Unknowns

The remainder when the number is divided by $36$ .

Constraints

The two remainders are different, so the number is not simply a common multiple plus one fixed remainder.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#2 Make a Systematic List

Instead of looking at what is left over, look at what is missing. Both remainders are exactly $2$ short of the divisor, so adding $2$ makes the number a common multiple of $9$ and $12$ .

3 · Execute4 carry out the plan

1Look at the shortfall

#16 Count the Complement 7.EE.A.2

A remainder of

7

out of

9

is

2

short of the next multiple of

9

; a remainder of

10

out of

12

is

2

short of the next multiple of

12

.

x = 9a + 7 \Rightarrow x + 2 = 9(a + 1), \quad x = 12b + 10 \Rightarrow x + 2 = 12(b + 1)

Adding

2

makes

x

a multiple of both

9

and

12

.

2Use the least common multiple

#13 Convert to Algebra 6.NS.B.4

A common multiple of

9

and

12

is a multiple of their least common multiple,

36

.

x + 2 = 36k

x

is

2

less than a multiple of

36

.

3Read off the remainder

#13 Convert to Algebra 7.EE.A.2

Rewrite

x

so that the part left over is between

0

and

35

.

x = 36k - 2 = 36(k - 1) + 34

Divided by

36

, the remainder is

34

.

4List a few to confirm

#2 Make a Systematic List 6.NS.B.4

The numbers are

34, 70, 106, \ldots

: each is

34

more than a multiple of

36

.

34 = 36 \times 0 + 34, \quad 70 = 36 \times 1 + 34

Every such number leaves

34

.

Answer: 34

4 · Reviewdoes it hold up?

Take $70$ : $70 = 9 \times 7 + 7$ and $70 = 12 \times 5 + 10$ , as required, and $70 = 36 \times 1 + 34$ .

Another way: List numbers that leave $10$ when divided by $12$ ( $10, 22, 34, \ldots$ ) and keep the first that leaves $7$ when divided by $9$ : that is $34$ , and adding $36$ keeps both remainders, so the remainder by $36$ is always $34$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the LCM of $9$ and $12$, which is $36$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting $x = 9a + 7$ as $x + 2 = 9(a + 1)$ and $36k - 2$ as $36(k - 1) + 34$.

💡Takeaway. When the remainders differ, check the shortfall -- if it is the same, add it back and use the LCM.

Variant 11 hard

A natural number leaves a remainder of $2$ when divided by $5$ , and a remainder of $5$ when divided by $8$ . What is the remainder when this number is divided by $40$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A number leaves $2$ when divided by $5$ and $5$ when divided by $8$ . Find what it leaves when divided by $40$ .

Givens

Divided by $5$ , the remainder is $2$ .
Divided by $8$ , the remainder is $5$ .

Unknowns

The remainder when the number is divided by $40$ .

Constraints

The two remainders are different, so the number is not simply a common multiple plus one fixed remainder.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#2 Make a Systematic List

Instead of looking at what is left over, look at what is missing. Both remainders are exactly $3$ short of the divisor, so adding $3$ makes the number a common multiple of $5$ and $8$ .

3 · Execute4 carry out the plan

1Look at the shortfall

#16 Count the Complement 7.EE.A.2

A remainder of

2

out of

5

is

3

short of the next multiple of

5

; a remainder of

5

out of

8

is

3

short of the next multiple of

8

.

x = 5a + 2 \Rightarrow x + 3 = 5(a + 1), \quad x = 8b + 5 \Rightarrow x + 3 = 8(b + 1)

Adding

3

makes

x

a multiple of both

5

and

8

.

2Use the least common multiple

#13 Convert to Algebra 6.NS.B.4

A common multiple of

5

and

8

is a multiple of their least common multiple,

40

.

x + 3 = 40k

x

is

3

less than a multiple of

40

.

3Read off the remainder

#13 Convert to Algebra 7.EE.A.2

Rewrite

x

so that the part left over is between

0

and

39

.

x = 40k - 3 = 40(k - 1) + 37

Divided by

40

, the remainder is

37

.

4List a few to confirm

#2 Make a Systematic List 6.NS.B.4

The numbers are

37, 77, 117, \ldots

: each is

37

more than a multiple of

40

.

37 = 40 \times 0 + 37, \quad 77 = 40 \times 1 + 37

Every such number leaves

37

.

Answer: 37

4 · Reviewdoes it hold up?

Take $77$ : $77 = 5 \times 15 + 2$ and $77 = 8 \times 9 + 5$ , as required, and $77 = 40 \times 1 + 37$ .

Another way: List numbers that leave $5$ when divided by $8$ ( $5, 13, 21, \ldots$ ) and keep the first that leaves $2$ when divided by $5$ : that is $37$ , and adding $40$ keeps both remainders, so the remainder by $40$ is always $37$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the LCM of $5$ and $8$, which is $40$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting $x = 5a + 2$ as $x + 3 = 5(a + 1)$ and $40k - 3$ as $40(k - 1) + 37$.

💡Takeaway. When the remainders differ, check the shortfall -- if it is the same, add it back and use the LCM.

Variant 12 hard

A natural number leaves a remainder of $4$ when divided by $7$ , and a remainder of $5$ when divided by $8$ . What is the remainder when this number is divided by $56$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A number leaves $4$ when divided by $7$ and $5$ when divided by $8$ . Find what it leaves when divided by $56$ .

Givens

Divided by $7$ , the remainder is $4$ .
Divided by $8$ , the remainder is $5$ .

Unknowns

The remainder when the number is divided by $56$ .

Constraints

The two remainders are different, so the number is not simply a common multiple plus one fixed remainder.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#2 Make a Systematic List

Instead of looking at what is left over, look at what is missing. Both remainders are exactly $3$ short of the divisor, so adding $3$ makes the number a common multiple of $7$ and $8$ .

3 · Execute4 carry out the plan

1Look at the shortfall

#16 Count the Complement 7.EE.A.2

A remainder of

4

out of

7

is

3

short of the next multiple of

7

; a remainder of

5

out of

8

is

3

short of the next multiple of

8

.

x = 7a + 4 \Rightarrow x + 3 = 7(a + 1), \quad x = 8b + 5 \Rightarrow x + 3 = 8(b + 1)

Adding

3

makes

x

a multiple of both

7

and

8

.

2Use the least common multiple

#13 Convert to Algebra 6.NS.B.4

A common multiple of

7

and

8

is a multiple of their least common multiple,

56

.

x + 3 = 56k

x

is

3

less than a multiple of

56

.

3Read off the remainder

#13 Convert to Algebra 7.EE.A.2

Rewrite

x

so that the part left over is between

0

and

55

.

x = 56k - 3 = 56(k - 1) + 53

Divided by

56

, the remainder is

53

.

4List a few to confirm

#2 Make a Systematic List 6.NS.B.4

The numbers are

53, 109, 165, \ldots

: each is

53

more than a multiple of

56

.

53 = 56 \times 0 + 53, \quad 109 = 56 \times 1 + 53

Every such number leaves

53

.

Answer: 53

4 · Reviewdoes it hold up?

Take $109$ : $109 = 7 \times 15 + 4$ and $109 = 8 \times 13 + 5$ , as required, and $109 = 56 \times 1 + 53$ .

Another way: List numbers that leave $5$ when divided by $8$ ( $5, 13, 21, \ldots$ ) and keep the first that leaves $4$ when divided by $7$ : that is $53$ , and adding $56$ keeps both remainders, so the remainder by $56$ is always $53$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Using the LCM of $7$ and $8$, which is $56$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting $x = 7a + 4$ as $x + 3 = 7(a + 1)$ and $56k - 3$ as $56(k - 1) + 53$.

💡Takeaway. When the remainders differ, check the shortfall -- if it is the same, add it back and use the LCM.

Same shortfall turns remainders into one LCM · 12 practice problems

Generated variants — 12

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