← A multiple is divisible by its factor · Divisibility and Remainder Reasoning

A multiple is divisible by its factor · 12 practice problems

4.OA.B.4

Generated variants — 12

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

Variant 1 easy answer: 5

The two numbers \bigodot and 2424 are in a factor-multiple relationship. How many two-digit numbers can \bigodot be?

Show solution
1 · Understandwhat's really being asked

A hidden two-digit number and 24 are a factor and a multiple of each other, in one order or the other. We must count how many two-digit numbers could be the hidden one.

Givens
  • The hidden number and 24 are in a factor-multiple relationship.
  • The hidden number has two digits, so it is from 10 to 99.
Unknowns
  • How many two-digit numbers fit.
Constraints
  • The relationship can go either way: the hidden number may be the factor or the multiple.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

'Factor-multiple relationship' does not say which one is which, so split it into two cases and list each. Listing beats guessing here because the two-digit range is small and a written list shows the repeats.

3 · Execute4 carry out the plan

1Split into two cases

#7 Identify Subproblems 4.OA.B.4
Either the hidden number divides 24 (it is a factor), or 24 divides it (it is a multiple). Both count, so handle each.
24or24\bigodot \mid 24\quad \text{or}\quad 24 \mid \bigodot
The phrase hides two questions; answering only one loses candidates.

2List the two-digit factors of 24

#2 Make a Systematic List 4.OA.B.4
Go through the factors of 24 and keep the ones with two digits.
12,2412, 24
These divide 24 exactly.

3List the two-digit multiples of 24

#2 Make a Systematic List 4.OA.B.4
Count up by 24 and keep the results that still have two digits.
24,48,72,9624, 48, 72, 96
These are what 24 divides exactly.

4Combine the lists and remove duplicates

#2 Make a Systematic List 4.OA.B.4
24 appears on both lists -- every number is a factor and a multiple of itself -- so count it once.
12,24,48,72,9612, 24, 48, 72, 96
That leaves 5 different two-digit numbers.
Answer: 5
4 · Reviewdoes it hold up?

Check the shared entry: 24 sits on both lists, and counting it twice would give 6 instead of 5.

Another way: Test every two-digit number in turn against 24, keeping any that divides it or is divided by it; the survivors are the same 5 numbers.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs for a whole number in the range 1-100 and recognize multiples — Listing the two-digit factors and the two-digit multiples of 24, then counting them without repeats.
💡Takeaway. 'Factor-multiple relationship' does not say which is which, so check both directions -- and count anything on both lists only once.
Variant 2 easy answer: 5

The two numbers \bigodot and 3030 are in a factor-multiple relationship. How many two-digit numbers can \bigodot be?

Show solution
1 · Understandwhat's really being asked

A hidden two-digit number and 30 are a factor and a multiple of each other, in one order or the other. We must count how many two-digit numbers could be the hidden one.

Givens
  • The hidden number and 30 are in a factor-multiple relationship.
  • The hidden number has two digits, so it is from 10 to 99.
Unknowns
  • How many two-digit numbers fit.
Constraints
  • The relationship can go either way: the hidden number may be the factor or the multiple.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

'Factor-multiple relationship' does not say which one is which, so split it into two cases and list each. Listing beats guessing here because the two-digit range is small and a written list shows the repeats.

3 · Execute4 carry out the plan

1Split into two cases

#7 Identify Subproblems 4.OA.B.4
Either the hidden number divides 30 (it is a factor), or 30 divides it (it is a multiple). Both count, so handle each.
30or30\bigodot \mid 30\quad \text{or}\quad 30 \mid \bigodot
The phrase hides two questions; answering only one loses candidates.

2List the two-digit factors of 30

#2 Make a Systematic List 4.OA.B.4
Go through the factors of 30 and keep the ones with two digits.
10,15,3010, 15, 30
These divide 30 exactly.

3List the two-digit multiples of 30

#2 Make a Systematic List 4.OA.B.4
Count up by 30 and keep the results that still have two digits.
30,60,9030, 60, 90
These are what 30 divides exactly.

4Combine the lists and remove duplicates

#2 Make a Systematic List 4.OA.B.4
30 appears on both lists -- every number is a factor and a multiple of itself -- so count it once.
10,15,30,60,9010, 15, 30, 60, 90
That leaves 5 different two-digit numbers.
Answer: 5
4 · Reviewdoes it hold up?

Check the shared entry: 30 sits on both lists, and counting it twice would give 6 instead of 5.

Another way: Test every two-digit number in turn against 30, keeping any that divides it or is divided by it; the survivors are the same 5 numbers.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs for a whole number in the range 1-100 and recognize multiples — Listing the two-digit factors and the two-digit multiples of 30, then counting them without repeats.
💡Takeaway. 'Factor-multiple relationship' does not say which is which, so check both directions -- and count anything on both lists only once.
Variant 3 easy answer: 4

The two numbers \bigodot and 3636 are in a factor-multiple relationship. How many two-digit numbers can \bigodot be?

Show solution
1 · Understandwhat's really being asked

A hidden two-digit number and 36 are a factor and a multiple of each other, in one order or the other. We must count how many two-digit numbers could be the hidden one.

Givens
  • The hidden number and 36 are in a factor-multiple relationship.
  • The hidden number has two digits, so it is from 10 to 99.
Unknowns
  • How many two-digit numbers fit.
Constraints
  • The relationship can go either way: the hidden number may be the factor or the multiple.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

'Factor-multiple relationship' does not say which one is which, so split it into two cases and list each. Listing beats guessing here because the two-digit range is small and a written list shows the repeats.

3 · Execute4 carry out the plan

1Split into two cases

#7 Identify Subproblems 4.OA.B.4
Either the hidden number divides 36 (it is a factor), or 36 divides it (it is a multiple). Both count, so handle each.
36or36\bigodot \mid 36\quad \text{or}\quad 36 \mid \bigodot
The phrase hides two questions; answering only one loses candidates.

2List the two-digit factors of 36

#2 Make a Systematic List 4.OA.B.4
Go through the factors of 36 and keep the ones with two digits.
12,18,3612, 18, 36
These divide 36 exactly.

3List the two-digit multiples of 36

#2 Make a Systematic List 4.OA.B.4
Count up by 36 and keep the results that still have two digits.
36,7236, 72
These are what 36 divides exactly.

4Combine the lists and remove duplicates

#2 Make a Systematic List 4.OA.B.4
36 appears on both lists -- every number is a factor and a multiple of itself -- so count it once.
12,18,36,7212, 18, 36, 72
That leaves 4 different two-digit numbers.
Answer: 4
4 · Reviewdoes it hold up?

Check the shared entry: 36 sits on both lists, and counting it twice would give 5 instead of 4.

Another way: Test every two-digit number in turn against 36, keeping any that divides it or is divided by it; the survivors are the same 4 numbers.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs for a whole number in the range 1-100 and recognize multiples — Listing the two-digit factors and the two-digit multiples of 36, then counting them without repeats.
💡Takeaway. 'Factor-multiple relationship' does not say which is which, so check both directions -- and count anything on both lists only once.
Variant 4 easy answer: 4

The two numbers \bigodot and 4040 are in a factor-multiple relationship. How many two-digit numbers can \bigodot be?

Show solution
1 · Understandwhat's really being asked

A hidden two-digit number and 40 are a factor and a multiple of each other, in one order or the other. We must count how many two-digit numbers could be the hidden one.

Givens
  • The hidden number and 40 are in a factor-multiple relationship.
  • The hidden number has two digits, so it is from 10 to 99.
Unknowns
  • How many two-digit numbers fit.
Constraints
  • The relationship can go either way: the hidden number may be the factor or the multiple.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

'Factor-multiple relationship' does not say which one is which, so split it into two cases and list each. Listing beats guessing here because the two-digit range is small and a written list shows the repeats.

3 · Execute4 carry out the plan

1Split into two cases

#7 Identify Subproblems 4.OA.B.4
Either the hidden number divides 40 (it is a factor), or 40 divides it (it is a multiple). Both count, so handle each.
40or40\bigodot \mid 40\quad \text{or}\quad 40 \mid \bigodot
The phrase hides two questions; answering only one loses candidates.

2List the two-digit factors of 40

#2 Make a Systematic List 4.OA.B.4
Go through the factors of 40 and keep the ones with two digits.
10,20,4010, 20, 40
These divide 40 exactly.

3List the two-digit multiples of 40

#2 Make a Systematic List 4.OA.B.4
Count up by 40 and keep the results that still have two digits.
40,8040, 80
These are what 40 divides exactly.

4Combine the lists and remove duplicates

#2 Make a Systematic List 4.OA.B.4
40 appears on both lists -- every number is a factor and a multiple of itself -- so count it once.
10,20,40,8010, 20, 40, 80
That leaves 4 different two-digit numbers.
Answer: 4
4 · Reviewdoes it hold up?

Check the shared entry: 40 sits on both lists, and counting it twice would give 5 instead of 4.

Another way: Test every two-digit number in turn against 40, keeping any that divides it or is divided by it; the survivors are the same 4 numbers.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs for a whole number in the range 1-100 and recognize multiples — Listing the two-digit factors and the two-digit multiples of 40, then counting them without repeats.
💡Takeaway. 'Factor-multiple relationship' does not say which is which, so check both directions -- and count anything on both lists only once.
Variant 5 medium answer: 3

The two numbers \bigodot and 4545 are in a factor-multiple relationship. How many two-digit numbers can \bigodot be?

Show solution
1 · Understandwhat's really being asked

A hidden two-digit number and 45 are a factor and a multiple of each other, in one order or the other. We must count how many two-digit numbers could be the hidden one.

Givens
  • The hidden number and 45 are in a factor-multiple relationship.
  • The hidden number has two digits, so it is from 10 to 99.
Unknowns
  • How many two-digit numbers fit.
Constraints
  • The relationship can go either way: the hidden number may be the factor or the multiple.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

'Factor-multiple relationship' does not say which one is which, so split it into two cases and list each. Listing beats guessing here because the two-digit range is small and a written list shows the repeats.

3 · Execute4 carry out the plan

1Split into two cases

#7 Identify Subproblems 4.OA.B.4
Either the hidden number divides 45 (it is a factor), or 45 divides it (it is a multiple). Both count, so handle each.
45or45\bigodot \mid 45\quad \text{or}\quad 45 \mid \bigodot
The phrase hides two questions; answering only one loses candidates.

2List the two-digit factors of 45

#2 Make a Systematic List 4.OA.B.4
Go through the factors of 45 and keep the ones with two digits.
15,4515, 45
These divide 45 exactly.

3List the two-digit multiples of 45

#2 Make a Systematic List 4.OA.B.4
Count up by 45 and keep the results that still have two digits.
45,9045, 90
These are what 45 divides exactly.

4Combine the lists and remove duplicates

#2 Make a Systematic List 4.OA.B.4
45 appears on both lists -- every number is a factor and a multiple of itself -- so count it once.
15,45,9015, 45, 90
That leaves 3 different two-digit numbers.
Answer: 3
4 · Reviewdoes it hold up?

Check the shared entry: 45 sits on both lists, and counting it twice would give 4 instead of 3.

Another way: Test every two-digit number in turn against 45, keeping any that divides it or is divided by it; the survivors are the same 3 numbers.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs for a whole number in the range 1-100 and recognize multiples — Listing the two-digit factors and the two-digit multiples of 45, then counting them without repeats.
💡Takeaway. 'Factor-multiple relationship' does not say which is which, so check both directions -- and count anything on both lists only once.
Variant 6 medium answer: 5

The two numbers \bigodot and 4848 are in a factor-multiple relationship. How many two-digit numbers can \bigodot be?

Show solution
1 · Understandwhat's really being asked

A hidden two-digit number and 48 are a factor and a multiple of each other, in one order or the other. We must count how many two-digit numbers could be the hidden one.

Givens
  • The hidden number and 48 are in a factor-multiple relationship.
  • The hidden number has two digits, so it is from 10 to 99.
Unknowns
  • How many two-digit numbers fit.
Constraints
  • The relationship can go either way: the hidden number may be the factor or the multiple.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

'Factor-multiple relationship' does not say which one is which, so split it into two cases and list each. Listing beats guessing here because the two-digit range is small and a written list shows the repeats.

3 · Execute4 carry out the plan

1Split into two cases

#7 Identify Subproblems 4.OA.B.4
Either the hidden number divides 48 (it is a factor), or 48 divides it (it is a multiple). Both count, so handle each.
48or48\bigodot \mid 48\quad \text{or}\quad 48 \mid \bigodot
The phrase hides two questions; answering only one loses candidates.

2List the two-digit factors of 48

#2 Make a Systematic List 4.OA.B.4
Go through the factors of 48 and keep the ones with two digits.
12,16,24,4812, 16, 24, 48
These divide 48 exactly.

3List the two-digit multiples of 48

#2 Make a Systematic List 4.OA.B.4
Count up by 48 and keep the results that still have two digits.
48,9648, 96
These are what 48 divides exactly.

4Combine the lists and remove duplicates

#2 Make a Systematic List 4.OA.B.4
48 appears on both lists -- every number is a factor and a multiple of itself -- so count it once.
12,16,24,48,9612, 16, 24, 48, 96
That leaves 5 different two-digit numbers.
Answer: 5
4 · Reviewdoes it hold up?

Check the shared entry: 48 sits on both lists, and counting it twice would give 6 instead of 5.

Another way: Test every two-digit number in turn against 48, keeping any that divides it or is divided by it; the survivors are the same 5 numbers.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs for a whole number in the range 1-100 and recognize multiples — Listing the two-digit factors and the two-digit multiples of 48, then counting them without repeats.
💡Takeaway. 'Factor-multiple relationship' does not say which is which, so check both directions -- and count anything on both lists only once.
Variant 7 medium answer: 3

The two numbers \bigodot and 5454 are in a factor-multiple relationship. How many two-digit numbers can \bigodot be?

Show solution
1 · Understandwhat's really being asked

A hidden two-digit number and 54 are a factor and a multiple of each other, in one order or the other. We must count how many two-digit numbers could be the hidden one.

Givens
  • The hidden number and 54 are in a factor-multiple relationship.
  • The hidden number has two digits, so it is from 10 to 99.
Unknowns
  • How many two-digit numbers fit.
Constraints
  • The relationship can go either way: the hidden number may be the factor or the multiple.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

'Factor-multiple relationship' does not say which one is which, so split it into two cases and list each. Listing beats guessing here because the two-digit range is small and a written list shows the repeats.

3 · Execute4 carry out the plan

1Split into two cases

#7 Identify Subproblems 4.OA.B.4
Either the hidden number divides 54 (it is a factor), or 54 divides it (it is a multiple). Both count, so handle each.
54or54\bigodot \mid 54\quad \text{or}\quad 54 \mid \bigodot
The phrase hides two questions; answering only one loses candidates.

2List the two-digit factors of 54

#2 Make a Systematic List 4.OA.B.4
Go through the factors of 54 and keep the ones with two digits.
18,27,5418, 27, 54
These divide 54 exactly.

3List the two-digit multiples of 54

#2 Make a Systematic List 4.OA.B.4
Count up by 54 and keep the results that still have two digits.
5454
These are what 54 divides exactly.

4Combine the lists and remove duplicates

#2 Make a Systematic List 4.OA.B.4
54 appears on both lists -- every number is a factor and a multiple of itself -- so count it once.
18,27,5418, 27, 54
That leaves 3 different two-digit numbers.
Answer: 3
4 · Reviewdoes it hold up?

Check the shared entry: 54 sits on both lists, and counting it twice would give 4 instead of 3.

Another way: Test every two-digit number in turn against 54, keeping any that divides it or is divided by it; the survivors are the same 3 numbers.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs for a whole number in the range 1-100 and recognize multiples — Listing the two-digit factors and the two-digit multiples of 54, then counting them without repeats.
💡Takeaway. 'Factor-multiple relationship' does not say which is which, so check both directions -- and count anything on both lists only once.
Variant 8 medium answer: 6

The two numbers \bigodot and 6060 are in a factor-multiple relationship. How many two-digit numbers can \bigodot be?

Show solution
1 · Understandwhat's really being asked

A hidden two-digit number and 60 are a factor and a multiple of each other, in one order or the other. We must count how many two-digit numbers could be the hidden one.

Givens
  • The hidden number and 60 are in a factor-multiple relationship.
  • The hidden number has two digits, so it is from 10 to 99.
Unknowns
  • How many two-digit numbers fit.
Constraints
  • The relationship can go either way: the hidden number may be the factor or the multiple.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

'Factor-multiple relationship' does not say which one is which, so split it into two cases and list each. Listing beats guessing here because the two-digit range is small and a written list shows the repeats.

3 · Execute4 carry out the plan

1Split into two cases

#7 Identify Subproblems 4.OA.B.4
Either the hidden number divides 60 (it is a factor), or 60 divides it (it is a multiple). Both count, so handle each.
60or60\bigodot \mid 60\quad \text{or}\quad 60 \mid \bigodot
The phrase hides two questions; answering only one loses candidates.

2List the two-digit factors of 60

#2 Make a Systematic List 4.OA.B.4
Go through the factors of 60 and keep the ones with two digits.
10,12,15,20,30,6010, 12, 15, 20, 30, 60
These divide 60 exactly.

3List the two-digit multiples of 60

#2 Make a Systematic List 4.OA.B.4
Count up by 60 and keep the results that still have two digits.
6060
These are what 60 divides exactly.

4Combine the lists and remove duplicates

#2 Make a Systematic List 4.OA.B.4
60 appears on both lists -- every number is a factor and a multiple of itself -- so count it once.
10,12,15,20,30,6010, 12, 15, 20, 30, 60
That leaves 6 different two-digit numbers.
Answer: 6
4 · Reviewdoes it hold up?

Check the shared entry: 60 sits on both lists, and counting it twice would give 7 instead of 6.

Another way: Test every two-digit number in turn against 60, keeping any that divides it or is divided by it; the survivors are the same 6 numbers.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs for a whole number in the range 1-100 and recognize multiples — Listing the two-digit factors and the two-digit multiples of 60, then counting them without repeats.
💡Takeaway. 'Factor-multiple relationship' does not say which is which, so check both directions -- and count anything on both lists only once.
Variant 9 hard answer: 3

The two numbers \bigodot and 6464 are in a factor-multiple relationship. How many two-digit numbers can \bigodot be?

Show solution
1 · Understandwhat's really being asked

A hidden two-digit number and 64 are a factor and a multiple of each other, in one order or the other. We must count how many two-digit numbers could be the hidden one.

Givens
  • The hidden number and 64 are in a factor-multiple relationship.
  • The hidden number has two digits, so it is from 10 to 99.
Unknowns
  • How many two-digit numbers fit.
Constraints
  • The relationship can go either way: the hidden number may be the factor or the multiple.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

'Factor-multiple relationship' does not say which one is which, so split it into two cases and list each. Listing beats guessing here because the two-digit range is small and a written list shows the repeats.

3 · Execute4 carry out the plan

1Split into two cases

#7 Identify Subproblems 4.OA.B.4
Either the hidden number divides 64 (it is a factor), or 64 divides it (it is a multiple). Both count, so handle each.
64or64\bigodot \mid 64\quad \text{or}\quad 64 \mid \bigodot
The phrase hides two questions; answering only one loses candidates.

2List the two-digit factors of 64

#2 Make a Systematic List 4.OA.B.4
Go through the factors of 64 and keep the ones with two digits.
16,32,6416, 32, 64
These divide 64 exactly.

3List the two-digit multiples of 64

#2 Make a Systematic List 4.OA.B.4
Count up by 64 and keep the results that still have two digits.
6464
These are what 64 divides exactly.

4Combine the lists and remove duplicates

#2 Make a Systematic List 4.OA.B.4
64 appears on both lists -- every number is a factor and a multiple of itself -- so count it once.
16,32,6416, 32, 64
That leaves 3 different two-digit numbers.
Answer: 3
4 · Reviewdoes it hold up?

Check the shared entry: 64 sits on both lists, and counting it twice would give 4 instead of 3.

Another way: Test every two-digit number in turn against 64, keeping any that divides it or is divided by it; the survivors are the same 3 numbers.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs for a whole number in the range 1-100 and recognize multiples — Listing the two-digit factors and the two-digit multiples of 64, then counting them without repeats.
💡Takeaway. 'Factor-multiple relationship' does not say which is which, so check both directions -- and count anything on both lists only once.
Variant 10 hard answer: 5

The two numbers \bigodot and 7272 are in a factor-multiple relationship. How many two-digit numbers can \bigodot be?

Show solution
1 · Understandwhat's really being asked

A hidden two-digit number and 72 are a factor and a multiple of each other, in one order or the other. We must count how many two-digit numbers could be the hidden one.

Givens
  • The hidden number and 72 are in a factor-multiple relationship.
  • The hidden number has two digits, so it is from 10 to 99.
Unknowns
  • How many two-digit numbers fit.
Constraints
  • The relationship can go either way: the hidden number may be the factor or the multiple.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

'Factor-multiple relationship' does not say which one is which, so split it into two cases and list each. Listing beats guessing here because the two-digit range is small and a written list shows the repeats.

3 · Execute4 carry out the plan

1Split into two cases

#7 Identify Subproblems 4.OA.B.4
Either the hidden number divides 72 (it is a factor), or 72 divides it (it is a multiple). Both count, so handle each.
72or72\bigodot \mid 72\quad \text{or}\quad 72 \mid \bigodot
The phrase hides two questions; answering only one loses candidates.

2List the two-digit factors of 72

#2 Make a Systematic List 4.OA.B.4
Go through the factors of 72 and keep the ones with two digits.
12,18,24,36,7212, 18, 24, 36, 72
These divide 72 exactly.

3List the two-digit multiples of 72

#2 Make a Systematic List 4.OA.B.4
Count up by 72 and keep the results that still have two digits.
7272
These are what 72 divides exactly.

4Combine the lists and remove duplicates

#2 Make a Systematic List 4.OA.B.4
72 appears on both lists -- every number is a factor and a multiple of itself -- so count it once.
12,18,24,36,7212, 18, 24, 36, 72
That leaves 5 different two-digit numbers.
Answer: 5
4 · Reviewdoes it hold up?

Check the shared entry: 72 sits on both lists, and counting it twice would give 6 instead of 5.

Another way: Test every two-digit number in turn against 72, keeping any that divides it or is divided by it; the survivors are the same 5 numbers.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs for a whole number in the range 1-100 and recognize multiples — Listing the two-digit factors and the two-digit multiples of 72, then counting them without repeats.
💡Takeaway. 'Factor-multiple relationship' does not say which is which, so check both directions -- and count anything on both lists only once.
Variant 11 hard answer: 5

The two numbers \bigodot and 8080 are in a factor-multiple relationship. How many two-digit numbers can \bigodot be?

Show solution
1 · Understandwhat's really being asked

A hidden two-digit number and 80 are a factor and a multiple of each other, in one order or the other. We must count how many two-digit numbers could be the hidden one.

Givens
  • The hidden number and 80 are in a factor-multiple relationship.
  • The hidden number has two digits, so it is from 10 to 99.
Unknowns
  • How many two-digit numbers fit.
Constraints
  • The relationship can go either way: the hidden number may be the factor or the multiple.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

'Factor-multiple relationship' does not say which one is which, so split it into two cases and list each. Listing beats guessing here because the two-digit range is small and a written list shows the repeats.

3 · Execute4 carry out the plan

1Split into two cases

#7 Identify Subproblems 4.OA.B.4
Either the hidden number divides 80 (it is a factor), or 80 divides it (it is a multiple). Both count, so handle each.
80or80\bigodot \mid 80\quad \text{or}\quad 80 \mid \bigodot
The phrase hides two questions; answering only one loses candidates.

2List the two-digit factors of 80

#2 Make a Systematic List 4.OA.B.4
Go through the factors of 80 and keep the ones with two digits.
10,16,20,40,8010, 16, 20, 40, 80
These divide 80 exactly.

3List the two-digit multiples of 80

#2 Make a Systematic List 4.OA.B.4
Count up by 80 and keep the results that still have two digits.
8080
These are what 80 divides exactly.

4Combine the lists and remove duplicates

#2 Make a Systematic List 4.OA.B.4
80 appears on both lists -- every number is a factor and a multiple of itself -- so count it once.
10,16,20,40,8010, 16, 20, 40, 80
That leaves 5 different two-digit numbers.
Answer: 5
4 · Reviewdoes it hold up?

Check the shared entry: 80 sits on both lists, and counting it twice would give 6 instead of 5.

Another way: Test every two-digit number in turn against 80, keeping any that divides it or is divided by it; the survivors are the same 5 numbers.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs for a whole number in the range 1-100 and recognize multiples — Listing the two-digit factors and the two-digit multiples of 80, then counting them without repeats.
💡Takeaway. 'Factor-multiple relationship' does not say which is which, so check both directions -- and count anything on both lists only once.
Variant 12 hard answer: 6

The two numbers \bigodot and 9090 are in a factor-multiple relationship. How many two-digit numbers can \bigodot be?

Show solution
1 · Understandwhat's really being asked

A hidden two-digit number and 90 are a factor and a multiple of each other, in one order or the other. We must count how many two-digit numbers could be the hidden one.

Givens
  • The hidden number and 90 are in a factor-multiple relationship.
  • The hidden number has two digits, so it is from 10 to 99.
Unknowns
  • How many two-digit numbers fit.
Constraints
  • The relationship can go either way: the hidden number may be the factor or the multiple.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

'Factor-multiple relationship' does not say which one is which, so split it into two cases and list each. Listing beats guessing here because the two-digit range is small and a written list shows the repeats.

3 · Execute4 carry out the plan

1Split into two cases

#7 Identify Subproblems 4.OA.B.4
Either the hidden number divides 90 (it is a factor), or 90 divides it (it is a multiple). Both count, so handle each.
90or90\bigodot \mid 90\quad \text{or}\quad 90 \mid \bigodot
The phrase hides two questions; answering only one loses candidates.

2List the two-digit factors of 90

#2 Make a Systematic List 4.OA.B.4
Go through the factors of 90 and keep the ones with two digits.
10,15,18,30,45,9010, 15, 18, 30, 45, 90
These divide 90 exactly.

3List the two-digit multiples of 90

#2 Make a Systematic List 4.OA.B.4
Count up by 90 and keep the results that still have two digits.
9090
These are what 90 divides exactly.

4Combine the lists and remove duplicates

#2 Make a Systematic List 4.OA.B.4
90 appears on both lists -- every number is a factor and a multiple of itself -- so count it once.
10,15,18,30,45,9010, 15, 18, 30, 45, 90
That leaves 6 different two-digit numbers.
Answer: 6
4 · Reviewdoes it hold up?

Check the shared entry: 90 sits on both lists, and counting it twice would give 7 instead of 6.

Another way: Test every two-digit number in turn against 90, keeping any that divides it or is divided by it; the survivors are the same 6 numbers.

Standardsmin grade 4
  • 4.OA.B.4 Find all factor pairs for a whole number in the range 1-100 and recognize multiples — Listing the two-digit factors and the two-digit multiples of 90, then counting them without repeats.
💡Takeaway. 'Factor-multiple relationship' does not say which is which, so check both directions -- and count anything on both lists only once.