Practice · Equal remainders: divisor divides every difference · Sensim Math

Variant 1 easy

When each of the four numbers $2613$ , $2243$ , $1503$ , and $985$ is divided by the same natural number, all four remainders are equal. Find every possible divisor, together with the remainder for that divisor. (The remainder is not $0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four numbers leave the same nonzero remainder when divided by one natural number. Find all such divisors and the matching remainders.

Givens

The numbers are $2613$ , $2243$ , $1503$ , $985$ .
All four leave the same remainder when divided by the same natural number.
The remainder is not $0$ .

Unknowns

Every possible divisor and its remainder.

Constraints

The common remainder is unknown, so it has to be removed from the picture first.

2 · Planchoose the strategy

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

The unknown remainder is the obstacle. Subtracting two of the numbers makes it cancel, so the divisor must divide every difference. Find the common divisors of the differences, then compute the remainder for each and drop any that give $0$ .

3 · Execute4 carry out the plan

1Subtract to cancel the remainder

#13 Convert to Algebra 7.EE.A.1

If the divisor is

G

and the common remainder is

r

, each number is a multiple of

G

plus

r

. The difference of two numbers loses the

r

.

(Ga + r) - (Gb + r) = G(a - b)

G

divides the difference of any two of the numbers.

2Compute the differences

#16 Count the Complement 6.NS.B.4

Subtract neighboring numbers. Every other difference is a sum of these, so these three are enough.

2613 - 2243 = 370, \quad 2243 - 1503 = 740, \quad 1503 - 985 = 518

G

is a common divisor of

370

,

740

,

518

.

3Find the common divisors

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

Factor each difference. The greatest common factor is

2 \times 37 = 74

, so the common divisors are the divisors of

74

.

370 = 2 \times 5 \times 37, \quad 740 = 2^2 \times 5 \times 37, \quad 518 = 2 \times 7 \times 37

G

can be

1

,

2

,

37

, or

74

.

4Find each remainder

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

Divide

985

by each candidate. Dividing by

1

always leaves

0

, which is not allowed.

985 = 74 \times 13 + 23, \quad 985 = 37 \times 26 + 23, \quad 985 = 2 \times 492 + 1

Divisor

74

leaves

23

, divisor

37

leaves

23

, divisor

2

leaves

1

.

Answer: divisor 74, remainder 23; divisor 37, remainder 23; divisor 2, remainder 1

4 · Reviewdoes it hold up?

Check $74$ on the largest number: $2613 = 74 \times 35 + 23$ , and $2243 = 74 \times 30 + 23$ , $1503 = 74 \times 20 + 23$ . All four numbers are odd, so dividing by $2$ leaves $1$ each time.

Another way: Use the differences from the smallest number instead: $2613 - 985 = 1628$ , $2243 - 985 = 1258$ , $1503 - 985 = 518$ . Their greatest common factor is also $74$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the greatest common factor of the differences and its divisors.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Factoring $(Ga + r) - (Gb + r)$ as $G(a - b)$.

💡Takeaway. When remainders are equal, subtract -- the remainder vanishes and the divisor is left in every difference.

Variant 2 easy

When each of the four numbers $2675$ , $2333$ , $1687$ , and $1117$ is divided by the same natural number, all four remainders are equal. Find every possible divisor, together with the remainder for that divisor. (The remainder is not $0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four numbers leave the same nonzero remainder when divided by one natural number. Find all such divisors and the matching remainders.

Givens

The numbers are $2675$ , $2333$ , $1687$ , $1117$ .
All four leave the same remainder when divided by the same natural number.
The remainder is not $0$ .

Unknowns

Every possible divisor and its remainder.

Constraints

The common remainder is unknown, so it has to be removed from the picture first.

2 · Planchoose the strategy

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

The unknown remainder is the obstacle. Subtracting two of the numbers makes it cancel, so the divisor must divide every difference. Find the common divisors of the differences, then compute the remainder for each and drop any that give $0$ .

3 · Execute4 carry out the plan

1Subtract to cancel the remainder

#13 Convert to Algebra 7.EE.A.1

If the divisor is

G

and the common remainder is

r

, each number is a multiple of

G

plus

r

. The difference of two numbers loses the

r

.

(Ga + r) - (Gb + r) = G(a - b)

G

divides the difference of any two of the numbers.

2Compute the differences

#16 Count the Complement 6.NS.B.4

Subtract neighboring numbers. Every other difference is a sum of these, so these three are enough.

2675 - 2333 = 342, \quad 2333 - 1687 = 646, \quad 1687 - 1117 = 570

G

is a common divisor of

342

,

646

,

570

.

3Find the common divisors

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

Factor each difference. The greatest common factor is

2 \times 19 = 38

, so the common divisors are the divisors of

38

.

342 = 2 \times 3^2 \times 19, \quad 646 = 2 \times 17 \times 19, \quad 570 = 2 \times 3 \times 5 \times 19

G

can be

1

,

2

,

19

, or

38

.

4Find each remainder

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

Divide

1117

by each candidate. Dividing by

1

always leaves

0

, which is not allowed.

1117 = 38 \times 29 + 15, \quad 1117 = 19 \times 58 + 15, \quad 1117 = 2 \times 558 + 1

Divisor

38

leaves

15

, divisor

19

leaves

15

, divisor

2

leaves

1

.

Answer: divisor 38, remainder 15; divisor 19, remainder 15; divisor 2, remainder 1

4 · Reviewdoes it hold up?

Check $38$ on the largest number: $2675 = 38 \times 70 + 15$ , and $2333 = 38 \times 61 + 15$ , $1687 = 38 \times 44 + 15$ . All four numbers are odd, so dividing by $2$ leaves $1$ each time.

Another way: Use the differences from the smallest number instead: $2675 - 1117 = 1558$ , $2333 - 1117 = 1216$ , $1687 - 1117 = 570$ . Their greatest common factor is also $38$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the greatest common factor of the differences and its divisors.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Factoring $(Ga + r) - (Gb + r)$ as $G(a - b)$.

💡Takeaway. When remainders are equal, subtract -- the remainder vanishes and the divisor is left in every difference.

Variant 3 easy

When each of the four numbers $8618$ , $7088$ , $5048$ , and $1393$ is divided by the same natural number, all four remainders are equal. Find every possible divisor, together with the remainder for that divisor. (The remainder is not $0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four numbers leave the same nonzero remainder when divided by one natural number. Find all such divisors and the matching remainders.

Givens

The numbers are $8618$ , $7088$ , $5048$ , $1393$ .
All four leave the same remainder when divided by the same natural number.
The remainder is not $0$ .

Unknowns

Every possible divisor and its remainder.

Constraints

The common remainder is unknown, so it has to be removed from the picture first.

2 · Planchoose the strategy

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

The unknown remainder is the obstacle. Subtracting two of the numbers makes it cancel, so the divisor must divide every difference. Find the common divisors of the differences, then compute the remainder for each and drop any that give $0$ .

3 · Execute4 carry out the plan

1Subtract to cancel the remainder

#13 Convert to Algebra 7.EE.A.1

If the divisor is

G

and the common remainder is

r

, each number is a multiple of

G

plus

r

. The difference of two numbers loses the

r

.

(Ga + r) - (Gb + r) = G(a - b)

G

divides the difference of any two of the numbers.

2Compute the differences

#16 Count the Complement 6.NS.B.4

Subtract neighboring numbers. Every other difference is a sum of these, so these three are enough.

8618 - 7088 = 1530, \quad 7088 - 5048 = 2040, \quad 5048 - 1393 = 3655

G

is a common divisor of

1530

,

2040

,

3655

.

3Find the common divisors

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

Factor each difference. The greatest common factor is

5 \times 17 = 85

, so the common divisors are the divisors of

85

.

1530 = 2 \times 3^2 \times 5 \times 17, \quad 2040 = 2^3 \times 3 \times 5 \times 17, \quad 3655 = 5 \times 17 \times 43

G

can be

1

,

5

,

17

, or

85

.

4Find each remainder

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

Divide

1393

by each candidate. Dividing by

1

always leaves

0

, which is not allowed.

1393 = 85 \times 16 + 33, \quad 1393 = 17 \times 81 + 16, \quad 1393 = 5 \times 278 + 3

Divisor

85

leaves

33

, divisor

17

leaves

16

, divisor

5

leaves

3

.

Answer: divisor 85, remainder 33; divisor 17, remainder 16; divisor 5, remainder 3

4 · Reviewdoes it hold up?

Check $85$ on the largest number: $8618 = 85 \times 101 + 33$ , and $7088 = 85 \times 83 + 33$ , $5048 = 85 \times 59 + 33$ . Each number is a multiple of $85$ plus $33$ , and $85$ is a multiple of $5$ , so dividing by $5$ leaves $3$ each time.

Another way: Use the differences from the smallest number instead: $8618 - 1393 = 7225$ , $7088 - 1393 = 5695$ , $5048 - 1393 = 3655$ . Their greatest common factor is also $85$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the greatest common factor of the differences and its divisors.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Factoring $(Ga + r) - (Gb + r)$ as $G(a - b)$.

💡Takeaway. When remainders are equal, subtract -- the remainder vanishes and the divisor is left in every difference.

Variant 4 easy

When each of the four numbers $8692$ , $7468$ , $4663$ , and $2317$ is divided by the same natural number, all four remainders are equal. Find every possible divisor, together with the remainder for that divisor. (The remainder is not $0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four numbers leave the same nonzero remainder when divided by one natural number. Find all such divisors and the matching remainders.

Givens

The numbers are $8692$ , $7468$ , $4663$ , $2317$ .
All four leave the same remainder when divided by the same natural number.
The remainder is not $0$ .

Unknowns

Every possible divisor and its remainder.

Constraints

The common remainder is unknown, so it has to be removed from the picture first.

2 · Planchoose the strategy

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

The unknown remainder is the obstacle. Subtracting two of the numbers makes it cancel, so the divisor must divide every difference. Find the common divisors of the differences, then compute the remainder for each and drop any that give $0$ .

3 · Execute4 carry out the plan

1Subtract to cancel the remainder

#13 Convert to Algebra 7.EE.A.1

If the divisor is

G

and the common remainder is

r

, each number is a multiple of

G

plus

r

. The difference of two numbers loses the

r

.

(Ga + r) - (Gb + r) = G(a - b)

G

divides the difference of any two of the numbers.

2Compute the differences

#16 Count the Complement 6.NS.B.4

Subtract neighboring numbers. Every other difference is a sum of these, so these three are enough.

8692 - 7468 = 1224, \quad 7468 - 4663 = 2805, \quad 4663 - 2317 = 2346

G

is a common divisor of

1224

,

2805

,

2346

.

3Find the common divisors

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

Factor each difference. The greatest common factor is

3 \times 17 = 51

, so the common divisors are the divisors of

51

.

1224 = 2^3 \times 3^2 \times 17, \quad 2805 = 3 \times 5 \times 11 \times 17, \quad 2346 = 2 \times 3 \times 17 \times 23

G

can be

1

,

3

,

17

, or

51

.

4Find each remainder

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

Divide

2317

by each candidate. Dividing by

1

always leaves

0

, which is not allowed.

2317 = 51 \times 45 + 22, \quad 2317 = 17 \times 136 + 5, \quad 2317 = 3 \times 772 + 1

Divisor

51

leaves

22

, divisor

17

leaves

5

, divisor

3

leaves

1

.

Answer: divisor 51, remainder 22; divisor 17, remainder 5; divisor 3, remainder 1

4 · Reviewdoes it hold up?

Check $51$ on the largest number: $8692 = 51 \times 170 + 22$ , and $7468 = 51 \times 146 + 22$ , $4663 = 51 \times 91 + 22$ . Each number is a multiple of $51$ plus $22$ , and $51$ is a multiple of $3$ , so dividing by $3$ leaves $1$ each time.

Another way: Use the differences from the smallest number instead: $8692 - 2317 = 6375$ , $7468 - 2317 = 5151$ , $4663 - 2317 = 2346$ . Their greatest common factor is also $51$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the greatest common factor of the differences and its divisors.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Factoring $(Ga + r) - (Gb + r)$ as $G(a - b)$.

💡Takeaway. When remainders are equal, subtract -- the remainder vanishes and the divisor is left in every difference.

Variant 5 medium

When each of the four numbers $8751$ , $7003$ , $4657$ , and $1851$ is divided by the same natural number, all four remainders are equal. Find every possible divisor, together with the remainder for that divisor. (The remainder is not $0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four numbers leave the same nonzero remainder when divided by one natural number. Find all such divisors and the matching remainders.

Givens

The numbers are $8751$ , $7003$ , $4657$ , $1851$ .
All four leave the same remainder when divided by the same natural number.
The remainder is not $0$ .

Unknowns

Every possible divisor and its remainder.

Constraints

The common remainder is unknown, so it has to be removed from the picture first.

2 · Planchoose the strategy

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

The unknown remainder is the obstacle. Subtracting two of the numbers makes it cancel, so the divisor must divide every difference. Find the common divisors of the differences, then compute the remainder for each and drop any that give $0$ .

3 · Execute4 carry out the plan

1Subtract to cancel the remainder

#13 Convert to Algebra 7.EE.A.1

If the divisor is

G

and the common remainder is

r

, each number is a multiple of

G

plus

r

. The difference of two numbers loses the

r

.

(Ga + r) - (Gb + r) = G(a - b)

G

divides the difference of any two of the numbers.

2Compute the differences

#16 Count the Complement 6.NS.B.4

Subtract neighboring numbers. Every other difference is a sum of these, so these three are enough.

8751 - 7003 = 1748, \quad 7003 - 4657 = 2346, \quad 4657 - 1851 = 2806

G

is a common divisor of

1748

,

2346

,

2806

.

3Find the common divisors

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

Factor each difference. The greatest common factor is

2 \times 23 = 46

, so the common divisors are the divisors of

46

.

1748 = 2^2 \times 19 \times 23, \quad 2346 = 2 \times 3 \times 17 \times 23, \quad 2806 = 2 \times 23 \times 61

G

can be

1

,

2

,

23

, or

46

.

4Find each remainder

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

Divide

1851

by each candidate. Dividing by

1

always leaves

0

, which is not allowed.

1851 = 46 \times 40 + 11, \quad 1851 = 23 \times 80 + 11, \quad 1851 = 2 \times 925 + 1

Divisor

46

leaves

11

, divisor

23

leaves

11

, divisor

2

leaves

1

.

Answer: divisor 46, remainder 11; divisor 23, remainder 11; divisor 2, remainder 1

4 · Reviewdoes it hold up?

Check $46$ on the largest number: $8751 = 46 \times 190 + 11$ , and $7003 = 46 \times 152 + 11$ , $4657 = 46 \times 101 + 11$ . All four numbers are odd, so dividing by $2$ leaves $1$ each time.

Another way: Use the differences from the smallest number instead: $8751 - 1851 = 6900$ , $7003 - 1851 = 5152$ , $4657 - 1851 = 2806$ . Their greatest common factor is also $46$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the greatest common factor of the differences and its divisors.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Factoring $(Ga + r) - (Gb + r)$ as $G(a - b)$.

💡Takeaway. When remainders are equal, subtract -- the remainder vanishes and the divisor is left in every difference.

Variant 6 medium

When each of the four numbers $8759$ , $7394$ , $5014$ , and $2879$ is divided by the same natural number, all four remainders are equal. Find every possible divisor, together with the remainder for that divisor. (The remainder is not $0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four numbers leave the same nonzero remainder when divided by one natural number. Find all such divisors and the matching remainders.

Givens

The numbers are $8759$ , $7394$ , $5014$ , $2879$ .
All four leave the same remainder when divided by the same natural number.
The remainder is not $0$ .

Unknowns

Every possible divisor and its remainder.

Constraints

The common remainder is unknown, so it has to be removed from the picture first.

2 · Planchoose the strategy

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

The unknown remainder is the obstacle. Subtracting two of the numbers makes it cancel, so the divisor must divide every difference. Find the common divisors of the differences, then compute the remainder for each and drop any that give $0$ .

3 · Execute4 carry out the plan

1Subtract to cancel the remainder

#13 Convert to Algebra 7.EE.A.1

If the divisor is

G

and the common remainder is

r

, each number is a multiple of

G

plus

r

. The difference of two numbers loses the

r

.

(Ga + r) - (Gb + r) = G(a - b)

G

divides the difference of any two of the numbers.

2Compute the differences

#16 Count the Complement 6.NS.B.4

Subtract neighboring numbers. Every other difference is a sum of these, so these three are enough.

8759 - 7394 = 1365, \quad 7394 - 5014 = 2380, \quad 5014 - 2879 = 2135

G

is a common divisor of

1365

,

2380

,

2135

.

3Find the common divisors

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

Factor each difference. The greatest common factor is

5 \times 7 = 35

, so the common divisors are the divisors of

35

.

1365 = 3 \times 5 \times 7 \times 13, \quad 2380 = 2^2 \times 5 \times 7 \times 17, \quad 2135 = 5 \times 7 \times 61

G

can be

1

,

5

,

7

, or

35

.

4Find each remainder

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

Divide

2879

by each candidate. Dividing by

1

always leaves

0

, which is not allowed.

2879 = 35 \times 82 + 9, \quad 2879 = 7 \times 411 + 2, \quad 2879 = 5 \times 575 + 4

Divisor

35

leaves

9

, divisor

7

leaves

2

, divisor

5

leaves

4

.

Answer: divisor 35, remainder 9; divisor 7, remainder 2; divisor 5, remainder 4

4 · Reviewdoes it hold up?

Check $35$ on the largest number: $8759 = 35 \times 250 + 9$ , and $7394 = 35 \times 211 + 9$ , $5014 = 35 \times 143 + 9$ . Each number is a multiple of $35$ plus $9$ , and $35$ is a multiple of $5$ , so dividing by $5$ leaves $4$ each time.

Another way: Use the differences from the smallest number instead: $8759 - 2879 = 5880$ , $7394 - 2879 = 4515$ , $5014 - 2879 = 2135$ . Their greatest common factor is also $35$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the greatest common factor of the differences and its divisors.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Factoring $(Ga + r) - (Gb + r)$ as $G(a - b)$.

💡Takeaway. When remainders are equal, subtract -- the remainder vanishes and the divisor is left in every difference.

Variant 7 medium

When each of the four numbers $8838$ , $7417$ , $4771$ , and $1929$ is divided by the same natural number, all four remainders are equal. Find every possible divisor, together with the remainder for that divisor. (The remainder is not $0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four numbers leave the same nonzero remainder when divided by one natural number. Find all such divisors and the matching remainders.

Givens

The numbers are $8838$ , $7417$ , $4771$ , $1929$ .
All four leave the same remainder when divided by the same natural number.
The remainder is not $0$ .

Unknowns

Every possible divisor and its remainder.

Constraints

The common remainder is unknown, so it has to be removed from the picture first.

2 · Planchoose the strategy

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

The unknown remainder is the obstacle. Subtracting two of the numbers makes it cancel, so the divisor must divide every difference. Find the common divisors of the differences, then compute the remainder for each and drop any that give $0$ .

3 · Execute4 carry out the plan

1Subtract to cancel the remainder

#13 Convert to Algebra 7.EE.A.1

If the divisor is

G

and the common remainder is

r

, each number is a multiple of

G

plus

r

. The difference of two numbers loses the

r

.

(Ga + r) - (Gb + r) = G(a - b)

G

divides the difference of any two of the numbers.

2Compute the differences

#16 Count the Complement 6.NS.B.4

Subtract neighboring numbers. Every other difference is a sum of these, so these three are enough.

8838 - 7417 = 1421, \quad 7417 - 4771 = 2646, \quad 4771 - 1929 = 2842

G

is a common divisor of

1421

,

2646

,

2842

.

3Find the common divisors

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

Factor each difference. The greatest common factor is

7^2 = 49

, so the common divisors are the divisors of

49

.

1421 = 7^2 \times 29, \quad 2646 = 2 \times 3^3 \times 7^2, \quad 2842 = 2 \times 7^2 \times 29

G

can be

1

,

7

, or

49

.

4Find each remainder

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

Divide

1929

by each candidate. Dividing by

1

always leaves

0

, which is not allowed.

1929 = 49 \times 39 + 18, \quad 1929 = 7 \times 275 + 4

Divisor

49

leaves

18

, divisor

7

leaves

4

.

Answer: divisor 49, remainder 18; divisor 7, remainder 4

4 · Reviewdoes it hold up?

Check $49$ on the largest number: $8838 = 49 \times 180 + 18$ , and $7417 = 49 \times 151 + 18$ , $4771 = 49 \times 97 + 18$ . Each number is a multiple of $49$ plus $18$ , and $49$ is a multiple of $7$ , so dividing by $7$ leaves $4$ each time.

Another way: Use the differences from the smallest number instead: $8838 - 1929 = 6909$ , $7417 - 1929 = 5488$ , $4771 - 1929 = 2842$ . Their greatest common factor is also $49$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the greatest common factor of the differences and its divisors.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Factoring $(Ga + r) - (Gb + r)$ as $G(a - b)$.

💡Takeaway. When remainders are equal, subtract -- the remainder vanishes and the divisor is left in every difference.

Variant 8 medium

When each of the four numbers $8983$ , $7534$ , $5326$ , and $2152$ is divided by the same natural number, all four remainders are equal. Find every possible divisor, together with the remainder for that divisor. (The remainder is not $0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four numbers leave the same nonzero remainder when divided by one natural number. Find all such divisors and the matching remainders.

Givens

The numbers are $8983$ , $7534$ , $5326$ , $2152$ .
All four leave the same remainder when divided by the same natural number.
The remainder is not $0$ .

Unknowns

Every possible divisor and its remainder.

Constraints

The common remainder is unknown, so it has to be removed from the picture first.

2 · Planchoose the strategy

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

The unknown remainder is the obstacle. Subtracting two of the numbers makes it cancel, so the divisor must divide every difference. Find the common divisors of the differences, then compute the remainder for each and drop any that give $0$ .

3 · Execute4 carry out the plan

1Subtract to cancel the remainder

#13 Convert to Algebra 7.EE.A.1

If the divisor is

G

and the common remainder is

r

, each number is a multiple of

G

plus

r

. The difference of two numbers loses the

r

.

(Ga + r) - (Gb + r) = G(a - b)

G

divides the difference of any two of the numbers.

2Compute the differences

#16 Count the Complement 6.NS.B.4

Subtract neighboring numbers. Every other difference is a sum of these, so these three are enough.

8983 - 7534 = 1449, \quad 7534 - 5326 = 2208, \quad 5326 - 2152 = 3174

G

is a common divisor of

1449

,

2208

,

3174

.

3Find the common divisors

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

Factor each difference. The greatest common factor is

3 \times 23 = 69

, so the common divisors are the divisors of

69

.

1449 = 3^2 \times 7 \times 23, \quad 2208 = 2^5 \times 3 \times 23, \quad 3174 = 2 \times 3 \times 23^2

G

can be

1

,

3

,

23

, or

69

.

4Find each remainder

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

Divide

2152

by each candidate. Dividing by

1

always leaves

0

, which is not allowed.

2152 = 69 \times 31 + 13, \quad 2152 = 23 \times 93 + 13, \quad 2152 = 3 \times 717 + 1

Divisor

69

leaves

13

, divisor

23

leaves

13

, divisor

3

leaves

1

.

Answer: divisor 69, remainder 13; divisor 23, remainder 13; divisor 3, remainder 1

4 · Reviewdoes it hold up?

Check $69$ on the largest number: $8983 = 69 \times 130 + 13$ , and $7534 = 69 \times 109 + 13$ , $5326 = 69 \times 77 + 13$ . Each number is a multiple of $69$ plus $13$ , and $69$ is a multiple of $3$ , so dividing by $3$ leaves $1$ each time.

Another way: Use the differences from the smallest number instead: $8983 - 2152 = 6831$ , $7534 - 2152 = 5382$ , $5326 - 2152 = 3174$ . Their greatest common factor is also $69$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the greatest common factor of the differences and its divisors.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Factoring $(Ga + r) - (Gb + r)$ as $G(a - b)$.

💡Takeaway. When remainders are equal, subtract -- the remainder vanishes and the divisor is left in every difference.

Variant 9 hard

When each of the four numbers $9017$ , $7405$ , $4615$ , and $1391$ is divided by the same natural number, all four remainders are equal. Find every possible divisor, together with the remainder for that divisor. (The remainder is not $0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four numbers leave the same nonzero remainder when divided by one natural number. Find all such divisors and the matching remainders.

Givens

The numbers are $9017$ , $7405$ , $4615$ , $1391$ .
All four leave the same remainder when divided by the same natural number.
The remainder is not $0$ .

Unknowns

Every possible divisor and its remainder.

Constraints

The common remainder is unknown, so it has to be removed from the picture first.

2 · Planchoose the strategy

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

The unknown remainder is the obstacle. Subtracting two of the numbers makes it cancel, so the divisor must divide every difference. Find the common divisors of the differences, then compute the remainder for each and drop any that give $0$ .

3 · Execute4 carry out the plan

1Subtract to cancel the remainder

#13 Convert to Algebra 7.EE.A.1

If the divisor is

G

and the common remainder is

r

, each number is a multiple of

G

plus

r

. The difference of two numbers loses the

r

.

(Ga + r) - (Gb + r) = G(a - b)

G

divides the difference of any two of the numbers.

2Compute the differences

#16 Count the Complement 6.NS.B.4

Subtract neighboring numbers. Every other difference is a sum of these, so these three are enough.

9017 - 7405 = 1612, \quad 7405 - 4615 = 2790, \quad 4615 - 1391 = 3224

G

is a common divisor of

1612

,

2790

,

3224

.

3Find the common divisors

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

Factor each difference. The greatest common factor is

2 \times 31 = 62

, so the common divisors are the divisors of

62

.

1612 = 2^2 \times 13 \times 31, \quad 2790 = 2 \times 3^2 \times 5 \times 31, \quad 3224 = 2^3 \times 13 \times 31

G

can be

1

,

2

,

31

, or

62

.

4Find each remainder

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

Divide

1391

by each candidate. Dividing by

1

always leaves

0

, which is not allowed.

1391 = 62 \times 22 + 27, \quad 1391 = 31 \times 44 + 27, \quad 1391 = 2 \times 695 + 1

Divisor

62

leaves

27

, divisor

31

leaves

27

, divisor

2

leaves

1

.

Answer: divisor 62, remainder 27; divisor 31, remainder 27; divisor 2, remainder 1

4 · Reviewdoes it hold up?

Check $62$ on the largest number: $9017 = 62 \times 145 + 27$ , and $7405 = 62 \times 119 + 27$ , $4615 = 62 \times 74 + 27$ . All four numbers are odd, so dividing by $2$ leaves $1$ each time.

Another way: Use the differences from the smallest number instead: $9017 - 1391 = 7626$ , $7405 - 1391 = 6014$ , $4615 - 1391 = 3224$ . Their greatest common factor is also $62$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the greatest common factor of the differences and its divisors.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Factoring $(Ga + r) - (Gb + r)$ as $G(a - b)$.

💡Takeaway. When remainders are equal, subtract -- the remainder vanishes and the divisor is left in every difference.

Variant 10 hard

When each of the four numbers $9041$ , $7819$ , $4623$ , and $2367$ is divided by the same natural number, all four remainders are equal. Find every possible divisor, together with the remainder for that divisor. (The remainder is not $0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four numbers leave the same nonzero remainder when divided by one natural number. Find all such divisors and the matching remainders.

Givens

The numbers are $9041$ , $7819$ , $4623$ , $2367$ .
All four leave the same remainder when divided by the same natural number.
The remainder is not $0$ .

Unknowns

Every possible divisor and its remainder.

Constraints

The common remainder is unknown, so it has to be removed from the picture first.

2 · Planchoose the strategy

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

The unknown remainder is the obstacle. Subtracting two of the numbers makes it cancel, so the divisor must divide every difference. Find the common divisors of the differences, then compute the remainder for each and drop any that give $0$ .

3 · Execute4 carry out the plan

1Subtract to cancel the remainder

#13 Convert to Algebra 7.EE.A.1

If the divisor is

G

and the common remainder is

r

, each number is a multiple of

G

plus

r

. The difference of two numbers loses the

r

.

(Ga + r) - (Gb + r) = G(a - b)

G

divides the difference of any two of the numbers.

2Compute the differences

#16 Count the Complement 6.NS.B.4

Subtract neighboring numbers. Every other difference is a sum of these, so these three are enough.

9041 - 7819 = 1222, \quad 7819 - 4623 = 3196, \quad 4623 - 2367 = 2256

G

is a common divisor of

1222

,

3196

,

2256

.

3Find the common divisors

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

Factor each difference. The greatest common factor is

2 \times 47 = 94

, so the common divisors are the divisors of

94

.

1222 = 2 \times 13 \times 47, \quad 3196 = 2^2 \times 17 \times 47, \quad 2256 = 2^4 \times 3 \times 47

G

can be

1

,

2

,

47

, or

94

.

4Find each remainder

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

Divide

2367

by each candidate. Dividing by

1

always leaves

0

, which is not allowed.

2367 = 94 \times 25 + 17, \quad 2367 = 47 \times 50 + 17, \quad 2367 = 2 \times 1183 + 1

Divisor

94

leaves

17

, divisor

47

leaves

17

, divisor

2

leaves

1

.

Answer: divisor 94, remainder 17; divisor 47, remainder 17; divisor 2, remainder 1

4 · Reviewdoes it hold up?

Check $94$ on the largest number: $9041 = 94 \times 96 + 17$ , and $7819 = 94 \times 83 + 17$ , $4623 = 94 \times 49 + 17$ . All four numbers are odd, so dividing by $2$ leaves $1$ each time.

Another way: Use the differences from the smallest number instead: $9041 - 2367 = 6674$ , $7819 - 2367 = 5452$ , $4623 - 2367 = 2256$ . Their greatest common factor is also $94$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the greatest common factor of the differences and its divisors.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Factoring $(Ga + r) - (Gb + r)$ as $G(a - b)$.

💡Takeaway. When remainders are equal, subtract -- the remainder vanishes and the divisor is left in every difference.

Variant 11 hard

When each of the four numbers $9197$ , $7487$ , $5036$ , and $2471$ is divided by the same natural number, all four remainders are equal. Find every possible divisor, together with the remainder for that divisor. (The remainder is not $0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four numbers leave the same nonzero remainder when divided by one natural number. Find all such divisors and the matching remainders.

Givens

The numbers are $9197$ , $7487$ , $5036$ , $2471$ .
All four leave the same remainder when divided by the same natural number.
The remainder is not $0$ .

Unknowns

Every possible divisor and its remainder.

Constraints

The common remainder is unknown, so it has to be removed from the picture first.

2 · Planchoose the strategy

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

The unknown remainder is the obstacle. Subtracting two of the numbers makes it cancel, so the divisor must divide every difference. Find the common divisors of the differences, then compute the remainder for each and drop any that give $0$ .

3 · Execute4 carry out the plan

1Subtract to cancel the remainder

#13 Convert to Algebra 7.EE.A.1

If the divisor is

G

and the common remainder is

r

, each number is a multiple of

G

plus

r

. The difference of two numbers loses the

r

.

(Ga + r) - (Gb + r) = G(a - b)

G

divides the difference of any two of the numbers.

2Compute the differences

#16 Count the Complement 6.NS.B.4

Subtract neighboring numbers. Every other difference is a sum of these, so these three are enough.

9197 - 7487 = 1710, \quad 7487 - 5036 = 2451, \quad 5036 - 2471 = 2565

G

is a common divisor of

1710

,

2451

,

2565

.

3Find the common divisors

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

Factor each difference. The greatest common factor is

3 \times 19 = 57

, so the common divisors are the divisors of

57

.

1710 = 2 \times 3^2 \times 5 \times 19, \quad 2451 = 3 \times 19 \times 43, \quad 2565 = 3^3 \times 5 \times 19

G

can be

1

,

3

,

19

, or

57

.

4Find each remainder

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

Divide

2471

by each candidate. Dividing by

1

always leaves

0

, which is not allowed.

2471 = 57 \times 43 + 20, \quad 2471 = 19 \times 130 + 1, \quad 2471 = 3 \times 823 + 2

Divisor

57

leaves

20

, divisor

19

leaves

1

, divisor

3

leaves

2

.

Answer: divisor 57, remainder 20; divisor 19, remainder 1; divisor 3, remainder 2

4 · Reviewdoes it hold up?

Check $57$ on the largest number: $9197 = 57 \times 161 + 20$ , and $7487 = 57 \times 131 + 20$ , $5036 = 57 \times 88 + 20$ . Each number is a multiple of $57$ plus $20$ , and $57$ is a multiple of $3$ , so dividing by $3$ leaves $2$ each time.

Another way: Use the differences from the smallest number instead: $9197 - 2471 = 6726$ , $7487 - 2471 = 5016$ , $5036 - 2471 = 2565$ . Their greatest common factor is also $57$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the greatest common factor of the differences and its divisors.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Factoring $(Ga + r) - (Gb + r)$ as $G(a - b)$.

💡Takeaway. When remainders are equal, subtract -- the remainder vanishes and the divisor is left in every difference.

Variant 12 hard

When each of the four numbers $9301$ , $7735$ , $4603$ , and $1761$ is divided by the same natural number, all four remainders are equal. Find every possible divisor, together with the remainder for that divisor. (The remainder is not $0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four numbers leave the same nonzero remainder when divided by one natural number. Find all such divisors and the matching remainders.

Givens

The numbers are $9301$ , $7735$ , $4603$ , $1761$ .
All four leave the same remainder when divided by the same natural number.
The remainder is not $0$ .

Unknowns

Every possible divisor and its remainder.

Constraints

The common remainder is unknown, so it has to be removed from the picture first.

2 · Planchoose the strategy

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

The unknown remainder is the obstacle. Subtracting two of the numbers makes it cancel, so the divisor must divide every difference. Find the common divisors of the differences, then compute the remainder for each and drop any that give $0$ .

3 · Execute4 carry out the plan

1Subtract to cancel the remainder

#13 Convert to Algebra 7.EE.A.1

If the divisor is

G

and the common remainder is

r

, each number is a multiple of

G

plus

r

. The difference of two numbers loses the

r

.

(Ga + r) - (Gb + r) = G(a - b)

G

divides the difference of any two of the numbers.

2Compute the differences

#16 Count the Complement 6.NS.B.4

Subtract neighboring numbers. Every other difference is a sum of these, so these three are enough.

9301 - 7735 = 1566, \quad 7735 - 4603 = 3132, \quad 4603 - 1761 = 2842

G

is a common divisor of

1566

,

3132

,

2842

.

3Find the common divisors

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

Factor each difference. The greatest common factor is

2 \times 29 = 58

, so the common divisors are the divisors of

58

.

1566 = 2 \times 3^3 \times 29, \quad 3132 = 2^2 \times 3^3 \times 29, \quad 2842 = 2 \times 7^2 \times 29

G

can be

1

,

2

,

29

, or

58

.

4Find each remainder

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

Divide

1761

by each candidate. Dividing by

1

always leaves

0

, which is not allowed.

1761 = 58 \times 30 + 21, \quad 1761 = 29 \times 60 + 21, \quad 1761 = 2 \times 880 + 1

Divisor

58

leaves

21

, divisor

29

leaves

21

, divisor

2

leaves

1

.

Answer: divisor 58, remainder 21; divisor 29, remainder 21; divisor 2, remainder 1

4 · Reviewdoes it hold up?

Check $58$ on the largest number: $9301 = 58 \times 160 + 21$ , and $7735 = 58 \times 133 + 21$ , $4603 = 58 \times 79 + 21$ . All four numbers are odd, so dividing by $2$ leaves $1$ each time.

Another way: Use the differences from the smallest number instead: $9301 - 1761 = 7540$ , $7735 - 1761 = 5974$ , $4603 - 1761 = 2842$ . Their greatest common factor is also $58$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the greatest common factor of the differences and its divisors.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Factoring $(Ga + r) - (Gb + r)$ as $G(a - b)$.

💡Takeaway. When remainders are equal, subtract -- the remainder vanishes and the divisor is left in every difference.

Equal remainders: divisor divides every difference · 12 practice problems

Generated variants — 12

Givens

Unknowns

Constraints

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Unknowns

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Where next?