← Count equal-share ways using divisors · Divisibility and Remainder Reasoning

Count equal-share ways using divisors · 12 practice problems

4.OA.B.43.OA.B.6

Generated variants — 12

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 easy answer: 4 ways

You want to share 2424 candies equally, with none left over, among more than 55 students. In how many different ways can the candies be shared?

Show solution
1 · Understandwhat's really being asked

24 candies are split evenly among a group of more than 5 students, with nothing left over. We must count how many different group sizes make that possible.

Givens
  • There are 24 candies in total.
  • Everyone gets the same number, and none are left over.
  • The number of students is more than 5.
Unknowns
  • How many different group sizes work.
Constraints
  • Each student's share is a whole number of candies.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

'Equal shares with none left over' is the definition of a factor, so list the factors of 24 in pairs -- that way none is missed -- and then keep the ones big enough.

3 · Execute4 carry out the plan

1Turn 'equal shares, none left over' into divisors

#2 Make a Systematic List 3.OA.B.6
If 24 candies split evenly among some students, that number of students divides 24 exactly.
24÷=whole number24 \div \square = \text{whole number}
The question is really about factors, not about candy.

2List all factor pairs of 24

#2 Make a Systematic List 4.OA.B.4
Pairing them up guarantees the list is complete: work upwards until the two halves of a pair meet.
24=1×24,2×12,3×8,4×624 = 1 \times 24, 2 \times 12, 3 \times 8, 4 \times 6
The factors are 1, 2, 3, 4, 6, 8, 12, 24.

3Keep only group sizes greater than 5

#6 Guess and Check 4.OA.B.4
The group must be bigger than 5, which rules out 1, 2, 3, 4.
6,8,12,246, 8, 12, 24
That leaves 4 group sizes.

4Confirm each surviving split works

#6 Guess and Check 3.OA.B.6
Divide to see each one really does share out evenly.
24÷6=4,24÷8=3,24÷12=2,24÷24=124 \div 6 = 4,\quad 24 \div 8 = 3,\quad 24 \div 12 = 2,\quad 24 \div 24 = 1
All 4 work, so there are 4 ways.
Answer: 4 ways
4 · Reviewdoes it hold up?

24 has 8 factors in all, and 4 of them are 5 or less, leaving 4.

Another way: Test each number from 6 up to 24 and keep the ones that divide 24; slower, but it gives the same 4.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing every factor of 24 in pairs and filtering by size.
  • 3.OA.B.6 Understand division as an unknown-factor problem — Reading 'shares evenly with none left over' as exact division.
💡Takeaway. 'Shares evenly with none left over' is another way of saying 'is a factor' -- so list the factor pairs and nothing is missed.
Variant 2 easy answer: 3 ways

You want to share 3030 candies equally, with none left over, among more than 66 students. In how many different ways can the candies be shared?

Show solution
1 · Understandwhat's really being asked

30 candies are split evenly among a group of more than 6 students, with nothing left over. We must count how many different group sizes make that possible.

Givens
  • There are 30 candies in total.
  • Everyone gets the same number, and none are left over.
  • The number of students is more than 6.
Unknowns
  • How many different group sizes work.
Constraints
  • Each student's share is a whole number of candies.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

'Equal shares with none left over' is the definition of a factor, so list the factors of 30 in pairs -- that way none is missed -- and then keep the ones big enough.

3 · Execute4 carry out the plan

1Turn 'equal shares, none left over' into divisors

#2 Make a Systematic List 3.OA.B.6
If 30 candies split evenly among some students, that number of students divides 30 exactly.
30÷=whole number30 \div \square = \text{whole number}
The question is really about factors, not about candy.

2List all factor pairs of 30

#2 Make a Systematic List 4.OA.B.4
Pairing them up guarantees the list is complete: work upwards until the two halves of a pair meet.
30=1×30,2×15,3×10,5×630 = 1 \times 30, 2 \times 15, 3 \times 10, 5 \times 6
The factors are 1, 2, 3, 5, 6, 10, 15, 30.

3Keep only group sizes greater than 6

#6 Guess and Check 4.OA.B.4
The group must be bigger than 6, which rules out 1, 2, 3, 5, 6.
10,15,3010, 15, 30
That leaves 3 group sizes.

4Confirm each surviving split works

#6 Guess and Check 3.OA.B.6
Divide to see each one really does share out evenly.
30÷10=3,30÷15=2,30÷30=130 \div 10 = 3,\quad 30 \div 15 = 2,\quad 30 \div 30 = 1
All 3 work, so there are 3 ways.
Answer: 3 ways
4 · Reviewdoes it hold up?

30 has 8 factors in all, and 5 of them are 6 or less, leaving 3.

Another way: Test each number from 7 up to 30 and keep the ones that divide 30; slower, but it gives the same 3.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing every factor of 30 in pairs and filtering by size.
  • 3.OA.B.6 Understand division as an unknown-factor problem — Reading 'shares evenly with none left over' as exact division.
💡Takeaway. 'Shares evenly with none left over' is another way of saying 'is a factor' -- so list the factor pairs and nothing is missed.
Variant 3 easy answer: 4 ways

You want to share 3636 candies equally, with none left over, among more than 88 students. In how many different ways can the candies be shared?

Show solution
1 · Understandwhat's really being asked

36 candies are split evenly among a group of more than 8 students, with nothing left over. We must count how many different group sizes make that possible.

Givens
  • There are 36 candies in total.
  • Everyone gets the same number, and none are left over.
  • The number of students is more than 8.
Unknowns
  • How many different group sizes work.
Constraints
  • Each student's share is a whole number of candies.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

'Equal shares with none left over' is the definition of a factor, so list the factors of 36 in pairs -- that way none is missed -- and then keep the ones big enough.

3 · Execute4 carry out the plan

1Turn 'equal shares, none left over' into divisors

#2 Make a Systematic List 3.OA.B.6
If 36 candies split evenly among some students, that number of students divides 36 exactly.
36÷=whole number36 \div \square = \text{whole number}
The question is really about factors, not about candy.

2List all factor pairs of 36

#2 Make a Systematic List 4.OA.B.4
Pairing them up guarantees the list is complete: work upwards until the two halves of a pair meet.
36=1×36,2×18,3×12,4×9,6×636 = 1 \times 36, 2 \times 18, 3 \times 12, 4 \times 9, 6 \times 6
The factors are 1, 2, 3, 4, 6, 9, 12, 18, 36.

3Keep only group sizes greater than 8

#6 Guess and Check 4.OA.B.4
The group must be bigger than 8, which rules out 1, 2, 3, 4, 6.
9,12,18,369, 12, 18, 36
That leaves 4 group sizes.

4Confirm each surviving split works

#6 Guess and Check 3.OA.B.6
Divide to see each one really does share out evenly.
36÷9=4,36÷12=3,36÷18=2,36÷36=136 \div 9 = 4,\quad 36 \div 12 = 3,\quad 36 \div 18 = 2,\quad 36 \div 36 = 1
All 4 work, so there are 4 ways.
Answer: 4 ways
4 · Reviewdoes it hold up?

36 has 9 factors in all, and 5 of them are 8 or less, leaving 4.

Another way: Test each number from 9 up to 36 and keep the ones that divide 36; slower, but it gives the same 4.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing every factor of 36 in pairs and filtering by size.
  • 3.OA.B.6 Understand division as an unknown-factor problem — Reading 'shares evenly with none left over' as exact division.
💡Takeaway. 'Shares evenly with none left over' is another way of saying 'is a factor' -- so list the factor pairs and nothing is missed.
Variant 4 easy answer: 3 ways

You want to share 4242 candies equally, with none left over, among more than 1010 students. In how many different ways can the candies be shared?

Show solution
1 · Understandwhat's really being asked

42 candies are split evenly among a group of more than 10 students, with nothing left over. We must count how many different group sizes make that possible.

Givens
  • There are 42 candies in total.
  • Everyone gets the same number, and none are left over.
  • The number of students is more than 10.
Unknowns
  • How many different group sizes work.
Constraints
  • Each student's share is a whole number of candies.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

'Equal shares with none left over' is the definition of a factor, so list the factors of 42 in pairs -- that way none is missed -- and then keep the ones big enough.

3 · Execute4 carry out the plan

1Turn 'equal shares, none left over' into divisors

#2 Make a Systematic List 3.OA.B.6
If 42 candies split evenly among some students, that number of students divides 42 exactly.
42÷=whole number42 \div \square = \text{whole number}
The question is really about factors, not about candy.

2List all factor pairs of 42

#2 Make a Systematic List 4.OA.B.4
Pairing them up guarantees the list is complete: work upwards until the two halves of a pair meet.
42=1×42,2×21,3×14,6×742 = 1 \times 42, 2 \times 21, 3 \times 14, 6 \times 7
The factors are 1, 2, 3, 6, 7, 14, 21, 42.

3Keep only group sizes greater than 10

#6 Guess and Check 4.OA.B.4
The group must be bigger than 10, which rules out 1, 2, 3, 6, 7.
14,21,4214, 21, 42
That leaves 3 group sizes.

4Confirm each surviving split works

#6 Guess and Check 3.OA.B.6
Divide to see each one really does share out evenly.
42÷14=3,42÷21=2,42÷42=142 \div 14 = 3,\quad 42 \div 21 = 2,\quad 42 \div 42 = 1
All 3 work, so there are 3 ways.
Answer: 3 ways
4 · Reviewdoes it hold up?

42 has 8 factors in all, and 5 of them are 10 or less, leaving 3.

Another way: Test each number from 11 up to 42 and keep the ones that divide 42; slower, but it gives the same 3.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing every factor of 42 in pairs and filtering by size.
  • 3.OA.B.6 Understand division as an unknown-factor problem — Reading 'shares evenly with none left over' as exact division.
💡Takeaway. 'Shares evenly with none left over' is another way of saying 'is a factor' -- so list the factor pairs and nothing is missed.
Variant 5 medium answer: 3 ways

You want to share 4848 candies equally, with none left over, among more than 1515 students. In how many different ways can the candies be shared?

Show solution
1 · Understandwhat's really being asked

48 candies are split evenly among a group of more than 15 students, with nothing left over. We must count how many different group sizes make that possible.

Givens
  • There are 48 candies in total.
  • Everyone gets the same number, and none are left over.
  • The number of students is more than 15.
Unknowns
  • How many different group sizes work.
Constraints
  • Each student's share is a whole number of candies.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

'Equal shares with none left over' is the definition of a factor, so list the factors of 48 in pairs -- that way none is missed -- and then keep the ones big enough.

3 · Execute4 carry out the plan

1Turn 'equal shares, none left over' into divisors

#2 Make a Systematic List 3.OA.B.6
If 48 candies split evenly among some students, that number of students divides 48 exactly.
48÷=whole number48 \div \square = \text{whole number}
The question is really about factors, not about candy.

2List all factor pairs of 48

#2 Make a Systematic List 4.OA.B.4
Pairing them up guarantees the list is complete: work upwards until the two halves of a pair meet.
48=1×48,2×24,3×16,4×12,6×848 = 1 \times 48, 2 \times 24, 3 \times 16, 4 \times 12, 6 \times 8
The factors are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.

3Keep only group sizes greater than 15

#6 Guess and Check 4.OA.B.4
The group must be bigger than 15, which rules out 1, 2, 3, 4, 6, 8, 12.
16,24,4816, 24, 48
That leaves 3 group sizes.

4Confirm each surviving split works

#6 Guess and Check 3.OA.B.6
Divide to see each one really does share out evenly.
48÷16=3,48÷24=2,48÷48=148 \div 16 = 3,\quad 48 \div 24 = 2,\quad 48 \div 48 = 1
All 3 work, so there are 3 ways.
Answer: 3 ways
4 · Reviewdoes it hold up?

48 has 10 factors in all, and 7 of them are 15 or less, leaving 3.

Another way: Test each number from 16 up to 48 and keep the ones that divide 48; slower, but it gives the same 3.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing every factor of 48 in pairs and filtering by size.
  • 3.OA.B.6 Understand division as an unknown-factor problem — Reading 'shares evenly with none left over' as exact division.
💡Takeaway. 'Shares evenly with none left over' is another way of saying 'is a factor' -- so list the factor pairs and nothing is missed.
Variant 6 medium answer: 3 ways

You want to share 5454 candies equally, with none left over, among more than 99 students. In how many different ways can the candies be shared?

Show solution
1 · Understandwhat's really being asked

54 candies are split evenly among a group of more than 9 students, with nothing left over. We must count how many different group sizes make that possible.

Givens
  • There are 54 candies in total.
  • Everyone gets the same number, and none are left over.
  • The number of students is more than 9.
Unknowns
  • How many different group sizes work.
Constraints
  • Each student's share is a whole number of candies.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

'Equal shares with none left over' is the definition of a factor, so list the factors of 54 in pairs -- that way none is missed -- and then keep the ones big enough.

3 · Execute4 carry out the plan

1Turn 'equal shares, none left over' into divisors

#2 Make a Systematic List 3.OA.B.6
If 54 candies split evenly among some students, that number of students divides 54 exactly.
54÷=whole number54 \div \square = \text{whole number}
The question is really about factors, not about candy.

2List all factor pairs of 54

#2 Make a Systematic List 4.OA.B.4
Pairing them up guarantees the list is complete: work upwards until the two halves of a pair meet.
54=1×54,2×27,3×18,6×954 = 1 \times 54, 2 \times 27, 3 \times 18, 6 \times 9
The factors are 1, 2, 3, 6, 9, 18, 27, 54.

3Keep only group sizes greater than 9

#6 Guess and Check 4.OA.B.4
The group must be bigger than 9, which rules out 1, 2, 3, 6, 9.
18,27,5418, 27, 54
That leaves 3 group sizes.

4Confirm each surviving split works

#6 Guess and Check 3.OA.B.6
Divide to see each one really does share out evenly.
54÷18=3,54÷27=2,54÷54=154 \div 18 = 3,\quad 54 \div 27 = 2,\quad 54 \div 54 = 1
All 3 work, so there are 3 ways.
Answer: 3 ways
4 · Reviewdoes it hold up?

54 has 8 factors in all, and 5 of them are 9 or less, leaving 3.

Another way: Test each number from 10 up to 54 and keep the ones that divide 54; slower, but it gives the same 3.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing every factor of 54 in pairs and filtering by size.
  • 3.OA.B.6 Understand division as an unknown-factor problem — Reading 'shares evenly with none left over' as exact division.
💡Takeaway. 'Shares evenly with none left over' is another way of saying 'is a factor' -- so list the factor pairs and nothing is missed.
Variant 7 medium answer: 4 ways

You want to share 6060 candies equally, with none left over, among more than 1212 students. In how many different ways can the candies be shared?

Show solution
1 · Understandwhat's really being asked

60 candies are split evenly among a group of more than 12 students, with nothing left over. We must count how many different group sizes make that possible.

Givens
  • There are 60 candies in total.
  • Everyone gets the same number, and none are left over.
  • The number of students is more than 12.
Unknowns
  • How many different group sizes work.
Constraints
  • Each student's share is a whole number of candies.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

'Equal shares with none left over' is the definition of a factor, so list the factors of 60 in pairs -- that way none is missed -- and then keep the ones big enough.

3 · Execute4 carry out the plan

1Turn 'equal shares, none left over' into divisors

#2 Make a Systematic List 3.OA.B.6
If 60 candies split evenly among some students, that number of students divides 60 exactly.
60÷=whole number60 \div \square = \text{whole number}
The question is really about factors, not about candy.

2List all factor pairs of 60

#2 Make a Systematic List 4.OA.B.4
Pairing them up guarantees the list is complete: work upwards until the two halves of a pair meet.
60=1×60,2×30,3×20,4×15,5×12,6×1060 = 1 \times 60, 2 \times 30, 3 \times 20, 4 \times 15, 5 \times 12, 6 \times 10
The factors are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.

3Keep only group sizes greater than 12

#6 Guess and Check 4.OA.B.4
The group must be bigger than 12, which rules out 1, 2, 3, 4, 5, 6, 10, 12.
15,20,30,6015, 20, 30, 60
That leaves 4 group sizes.

4Confirm each surviving split works

#6 Guess and Check 3.OA.B.6
Divide to see each one really does share out evenly.
60÷15=4,60÷20=3,60÷30=2,60÷60=160 \div 15 = 4,\quad 60 \div 20 = 3,\quad 60 \div 30 = 2,\quad 60 \div 60 = 1
All 4 work, so there are 4 ways.
Answer: 4 ways
4 · Reviewdoes it hold up?

60 has 12 factors in all, and 8 of them are 12 or less, leaving 4.

Another way: Test each number from 13 up to 60 and keep the ones that divide 60; slower, but it gives the same 4.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing every factor of 60 in pairs and filtering by size.
  • 3.OA.B.6 Understand division as an unknown-factor problem — Reading 'shares evenly with none left over' as exact division.
💡Takeaway. 'Shares evenly with none left over' is another way of saying 'is a factor' -- so list the factor pairs and nothing is missed.
Variant 8 medium answer: 3 ways

You want to share 6666 candies equally, with none left over, among more than 1111 students. In how many different ways can the candies be shared?

Show solution
1 · Understandwhat's really being asked

66 candies are split evenly among a group of more than 11 students, with nothing left over. We must count how many different group sizes make that possible.

Givens
  • There are 66 candies in total.
  • Everyone gets the same number, and none are left over.
  • The number of students is more than 11.
Unknowns
  • How many different group sizes work.
Constraints
  • Each student's share is a whole number of candies.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

'Equal shares with none left over' is the definition of a factor, so list the factors of 66 in pairs -- that way none is missed -- and then keep the ones big enough.

3 · Execute4 carry out the plan

1Turn 'equal shares, none left over' into divisors

#2 Make a Systematic List 3.OA.B.6
If 66 candies split evenly among some students, that number of students divides 66 exactly.
66÷=whole number66 \div \square = \text{whole number}
The question is really about factors, not about candy.

2List all factor pairs of 66

#2 Make a Systematic List 4.OA.B.4
Pairing them up guarantees the list is complete: work upwards until the two halves of a pair meet.
66=1×66,2×33,3×22,6×1166 = 1 \times 66, 2 \times 33, 3 \times 22, 6 \times 11
The factors are 1, 2, 3, 6, 11, 22, 33, 66.

3Keep only group sizes greater than 11

#6 Guess and Check 4.OA.B.4
The group must be bigger than 11, which rules out 1, 2, 3, 6, 11.
22,33,6622, 33, 66
That leaves 3 group sizes.

4Confirm each surviving split works

#6 Guess and Check 3.OA.B.6
Divide to see each one really does share out evenly.
66÷22=3,66÷33=2,66÷66=166 \div 22 = 3,\quad 66 \div 33 = 2,\quad 66 \div 66 = 1
All 3 work, so there are 3 ways.
Answer: 3 ways
4 · Reviewdoes it hold up?

66 has 8 factors in all, and 5 of them are 11 or less, leaving 3.

Another way: Test each number from 12 up to 66 and keep the ones that divide 66; slower, but it gives the same 3.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing every factor of 66 in pairs and filtering by size.
  • 3.OA.B.6 Understand division as an unknown-factor problem — Reading 'shares evenly with none left over' as exact division.
💡Takeaway. 'Shares evenly with none left over' is another way of saying 'is a factor' -- so list the factor pairs and nothing is missed.
Variant 9 hard answer: 3 ways

You want to share 7272 candies equally, with none left over, among more than 2020 students. In how many different ways can the candies be shared?

Show solution
1 · Understandwhat's really being asked

72 candies are split evenly among a group of more than 20 students, with nothing left over. We must count how many different group sizes make that possible.

Givens
  • There are 72 candies in total.
  • Everyone gets the same number, and none are left over.
  • The number of students is more than 20.
Unknowns
  • How many different group sizes work.
Constraints
  • Each student's share is a whole number of candies.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

'Equal shares with none left over' is the definition of a factor, so list the factors of 72 in pairs -- that way none is missed -- and then keep the ones big enough.

3 · Execute4 carry out the plan

1Turn 'equal shares, none left over' into divisors

#2 Make a Systematic List 3.OA.B.6
If 72 candies split evenly among some students, that number of students divides 72 exactly.
72÷=whole number72 \div \square = \text{whole number}
The question is really about factors, not about candy.

2List all factor pairs of 72

#2 Make a Systematic List 4.OA.B.4
Pairing them up guarantees the list is complete: work upwards until the two halves of a pair meet.
72=1×72,2×36,3×24,4×18,6×12,8×972 = 1 \times 72, 2 \times 36, 3 \times 24, 4 \times 18, 6 \times 12, 8 \times 9
The factors are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.

3Keep only group sizes greater than 20

#6 Guess and Check 4.OA.B.4
The group must be bigger than 20, which rules out 1, 2, 3, 4, 6, 8, 9, 12, 18.
24,36,7224, 36, 72
That leaves 3 group sizes.

4Confirm each surviving split works

#6 Guess and Check 3.OA.B.6
Divide to see each one really does share out evenly.
72÷24=3,72÷36=2,72÷72=172 \div 24 = 3,\quad 72 \div 36 = 2,\quad 72 \div 72 = 1
All 3 work, so there are 3 ways.
Answer: 3 ways
4 · Reviewdoes it hold up?

72 has 12 factors in all, and 9 of them are 20 or less, leaving 3.

Another way: Test each number from 21 up to 72 and keep the ones that divide 72; slower, but it gives the same 3.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing every factor of 72 in pairs and filtering by size.
  • 3.OA.B.6 Understand division as an unknown-factor problem — Reading 'shares evenly with none left over' as exact division.
💡Takeaway. 'Shares evenly with none left over' is another way of saying 'is a factor' -- so list the factor pairs and nothing is missed.
Variant 10 hard answer: 4 ways

You want to share 8484 candies equally, with none left over, among more than 1818 students. In how many different ways can the candies be shared?

Show solution
1 · Understandwhat's really being asked

84 candies are split evenly among a group of more than 18 students, with nothing left over. We must count how many different group sizes make that possible.

Givens
  • There are 84 candies in total.
  • Everyone gets the same number, and none are left over.
  • The number of students is more than 18.
Unknowns
  • How many different group sizes work.
Constraints
  • Each student's share is a whole number of candies.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

'Equal shares with none left over' is the definition of a factor, so list the factors of 84 in pairs -- that way none is missed -- and then keep the ones big enough.

3 · Execute4 carry out the plan

1Turn 'equal shares, none left over' into divisors

#2 Make a Systematic List 3.OA.B.6
If 84 candies split evenly among some students, that number of students divides 84 exactly.
84÷=whole number84 \div \square = \text{whole number}
The question is really about factors, not about candy.

2List all factor pairs of 84

#2 Make a Systematic List 4.OA.B.4
Pairing them up guarantees the list is complete: work upwards until the two halves of a pair meet.
84=1×84,2×42,3×28,4×21,6×14,7×1284 = 1 \times 84, 2 \times 42, 3 \times 28, 4 \times 21, 6 \times 14, 7 \times 12
The factors are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84.

3Keep only group sizes greater than 18

#6 Guess and Check 4.OA.B.4
The group must be bigger than 18, which rules out 1, 2, 3, 4, 6, 7, 12, 14.
21,28,42,8421, 28, 42, 84
That leaves 4 group sizes.

4Confirm each surviving split works

#6 Guess and Check 3.OA.B.6
Divide to see each one really does share out evenly.
84÷21=4,84÷28=3,84÷42=2,84÷84=184 \div 21 = 4,\quad 84 \div 28 = 3,\quad 84 \div 42 = 2,\quad 84 \div 84 = 1
All 4 work, so there are 4 ways.
Answer: 4 ways
4 · Reviewdoes it hold up?

84 has 12 factors in all, and 8 of them are 18 or less, leaving 4.

Another way: Test each number from 19 up to 84 and keep the ones that divide 84; slower, but it gives the same 4.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing every factor of 84 in pairs and filtering by size.
  • 3.OA.B.6 Understand division as an unknown-factor problem — Reading 'shares evenly with none left over' as exact division.
💡Takeaway. 'Shares evenly with none left over' is another way of saying 'is a factor' -- so list the factor pairs and nothing is missed.
Variant 11 hard answer: 3 ways

You want to share 9090 candies equally, with none left over, among more than 2525 students. In how many different ways can the candies be shared?

Show solution
1 · Understandwhat's really being asked

90 candies are split evenly among a group of more than 25 students, with nothing left over. We must count how many different group sizes make that possible.

Givens
  • There are 90 candies in total.
  • Everyone gets the same number, and none are left over.
  • The number of students is more than 25.
Unknowns
  • How many different group sizes work.
Constraints
  • Each student's share is a whole number of candies.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

'Equal shares with none left over' is the definition of a factor, so list the factors of 90 in pairs -- that way none is missed -- and then keep the ones big enough.

3 · Execute4 carry out the plan

1Turn 'equal shares, none left over' into divisors

#2 Make a Systematic List 3.OA.B.6
If 90 candies split evenly among some students, that number of students divides 90 exactly.
90÷=whole number90 \div \square = \text{whole number}
The question is really about factors, not about candy.

2List all factor pairs of 90

#2 Make a Systematic List 4.OA.B.4
Pairing them up guarantees the list is complete: work upwards until the two halves of a pair meet.
90=1×90,2×45,3×30,5×18,6×15,9×1090 = 1 \times 90, 2 \times 45, 3 \times 30, 5 \times 18, 6 \times 15, 9 \times 10
The factors are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90.

3Keep only group sizes greater than 25

#6 Guess and Check 4.OA.B.4
The group must be bigger than 25, which rules out 1, 2, 3, 5, 6, 9, 10, 15, 18.
30,45,9030, 45, 90
That leaves 3 group sizes.

4Confirm each surviving split works

#6 Guess and Check 3.OA.B.6
Divide to see each one really does share out evenly.
90÷30=3,90÷45=2,90÷90=190 \div 30 = 3,\quad 90 \div 45 = 2,\quad 90 \div 90 = 1
All 3 work, so there are 3 ways.
Answer: 3 ways
4 · Reviewdoes it hold up?

90 has 12 factors in all, and 9 of them are 25 or less, leaving 3.

Another way: Test each number from 26 up to 90 and keep the ones that divide 90; slower, but it gives the same 3.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing every factor of 90 in pairs and filtering by size.
  • 3.OA.B.6 Understand division as an unknown-factor problem — Reading 'shares evenly with none left over' as exact division.
💡Takeaway. 'Shares evenly with none left over' is another way of saying 'is a factor' -- so list the factor pairs and nothing is missed.
Variant 12 hard answer: 3 ways

You want to share 9696 candies equally, with none left over, among more than 3030 students. In how many different ways can the candies be shared?

Show solution
1 · Understandwhat's really being asked

96 candies are split evenly among a group of more than 30 students, with nothing left over. We must count how many different group sizes make that possible.

Givens
  • There are 96 candies in total.
  • Everyone gets the same number, and none are left over.
  • The number of students is more than 30.
Unknowns
  • How many different group sizes work.
Constraints
  • Each student's share is a whole number of candies.
2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #6 Guess and Check

'Equal shares with none left over' is the definition of a factor, so list the factors of 96 in pairs -- that way none is missed -- and then keep the ones big enough.

3 · Execute4 carry out the plan

1Turn 'equal shares, none left over' into divisors

#2 Make a Systematic List 3.OA.B.6
If 96 candies split evenly among some students, that number of students divides 96 exactly.
96÷=whole number96 \div \square = \text{whole number}
The question is really about factors, not about candy.

2List all factor pairs of 96

#2 Make a Systematic List 4.OA.B.4
Pairing them up guarantees the list is complete: work upwards until the two halves of a pair meet.
96=1×96,2×48,3×32,4×24,6×16,8×1296 = 1 \times 96, 2 \times 48, 3 \times 32, 4 \times 24, 6 \times 16, 8 \times 12
The factors are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96.

3Keep only group sizes greater than 30

#6 Guess and Check 4.OA.B.4
The group must be bigger than 30, which rules out 1, 2, 3, 4, 6, 8, 12, 16, 24.
32,48,9632, 48, 96
That leaves 3 group sizes.

4Confirm each surviving split works

#6 Guess and Check 3.OA.B.6
Divide to see each one really does share out evenly.
96÷32=3,96÷48=2,96÷96=196 \div 32 = 3,\quad 96 \div 48 = 2,\quad 96 \div 96 = 1
All 3 work, so there are 3 ways.
Answer: 3 ways
4 · Reviewdoes it hold up?

96 has 12 factors in all, and 9 of them are 30 or less, leaving 3.

Another way: Test each number from 31 up to 96 and keep the ones that divide 96; slower, but it gives the same 3.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing every factor of 96 in pairs and filtering by size.
  • 3.OA.B.6 Understand division as an unknown-factor problem — Reading 'shares evenly with none left over' as exact division.
💡Takeaway. 'Shares evenly with none left over' is another way of saying 'is a factor' -- so list the factor pairs and nothing is missed.