← Different factor pairs, same product · Decompose a Number into Parts and Factors

Different factor pairs, same product · 12 practice problems

3.OA.B.54.OA.B.4

Generated variants — 12

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

Variant 1 easy answer: 3 triangles

A piece of wire is as long as a 6 cm6\ \text{cm} stick laid end to end 66 times. Using this wire, what is the greatest number of triangles you can make if each triangle has three sides that are all 4 cm4\ \text{cm} long?

Show solution
1 · Understandwhat's really being asked

A wire is as long as a 6 cm stick placed end to end 6 times. Each triangle is made of three 4 cm sides. We want the greatest number of such triangles the wire can make.

Givens
  • The wire length equals a 6 cm stick laid end to end 6 times.
  • Each triangle has three sides, each 4 cm long.
Unknowns
  • The greatest number of 4 cm equilateral triangles the wire can make.
Constraints
  • Each triangle uses exactly 3 sides of 4 cm.
  • Whole triangles only (you cannot use part of a triangle).
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units

Break the problem into small steps: first find the total wire length, then find how much wire one triangle needs, and finally see how many triangles fit. Tracking centimeters keeps the multiplication and division lined up correctly.

3 · Execute3 carry out the plan

1Find the total wire length

#7 Identify Subproblems 3.OA.B.5
The wire is a 6 cm length repeated 6 times, so multiply 6 by 6.
6×6=366 \times 6 = 36
6 equal lengths of 6 cm is just the multiplication fact 6 times 6.

2Find the wire one triangle needs

#8 Analyze the Units 3.OA.B.5
Each triangle has three 4 cm sides, so one triangle uses 4 cm three times.
4×3=124 \times 3 = 12
Three sides of the same length is simply 4 added three times, an easy times-three fact.

3Count how many triangles fit

#7 Identify Subproblems 4.OA.B.4
See how many 12 cm pieces fit in 36 cm. Because 12 times 3 equals 36, exactly 3 triangles can be made with no wire left over.
36÷12=336 \div 12 = 3
Finding how many equal groups of 12 make 36 is the same as knowing the factor pair 12 and 3 of 36.
Answer: 3 triangles
4 · Reviewdoes it hold up?

3 triangles use 12 times 3 = 36 cm, which is exactly the whole wire, so the answer fits with nothing wasted and the magnitude is sensible.

Another way: You could repeatedly subtract 12 cm from 36 cm and count the 3 subtractions, which is the repeated-subtraction view of the same division.

Standardsmin grade 4
  • 3.OA.B.5 Apply properties of operations as strategies to multiply and divide — Multiplying 6 by 6 for the total wire length and 4 by 3 for one triangle's wire.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that 36 splits into equal groups of 12, giving 3 triangles.
💡Takeaway. Find the whole length, find one triangle's length, then split into equal groups -- only Grade 4 grouping you already know!
Variant 2 easy answer: 4 triangles

A piece of wire is as long as a 6 cm6\ \text{cm} stick laid end to end 44 times. Using this wire, what is the greatest number of triangles you can make if each triangle has three sides that are all 2 cm2\ \text{cm} long?

Show solution
1 · Understandwhat's really being asked

A wire is as long as a 6 cm stick placed end to end 4 times. Each triangle is made of three 2 cm sides. We want the greatest number of such triangles the wire can make.

Givens
  • The wire length equals a 6 cm stick laid end to end 4 times.
  • Each triangle has three sides, each 2 cm long.
Unknowns
  • The greatest number of 2 cm equilateral triangles the wire can make.
Constraints
  • Each triangle uses exactly 3 sides of 2 cm.
  • Whole triangles only (you cannot use part of a triangle).
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units

Break the problem into small steps: first find the total wire length, then find how much wire one triangle needs, and finally see how many triangles fit. Tracking centimeters keeps the multiplication and division lined up correctly.

3 · Execute3 carry out the plan

1Find the total wire length

#7 Identify Subproblems 3.OA.B.5
The wire is a 6 cm length repeated 4 times, so multiply 6 by 4.
6×4=246 \times 4 = 24
4 equal lengths of 6 cm is just the multiplication fact 6 times 4.

2Find the wire one triangle needs

#8 Analyze the Units 3.OA.B.5
Each triangle has three 2 cm sides, so one triangle uses 2 cm three times.
2×3=62 \times 3 = 6
Three sides of the same length is simply 2 added three times, an easy times-three fact.

3Count how many triangles fit

#7 Identify Subproblems 4.OA.B.4
See how many 6 cm pieces fit in 24 cm. Because 6 times 4 equals 24, exactly 4 triangles can be made with no wire left over.
24÷6=424 \div 6 = 4
Finding how many equal groups of 6 make 24 is the same as knowing the factor pair 6 and 4 of 24.
Answer: 4 triangles
4 · Reviewdoes it hold up?

4 triangles use 6 times 4 = 24 cm, which is exactly the whole wire, so the answer fits with nothing wasted and the magnitude is sensible.

Another way: You could repeatedly subtract 6 cm from 24 cm and count the 4 subtractions, which is the repeated-subtraction view of the same division.

Standardsmin grade 4
  • 3.OA.B.5 Apply properties of operations as strategies to multiply and divide — Multiplying 6 by 4 for the total wire length and 2 by 3 for one triangle's wire.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that 24 splits into equal groups of 6, giving 4 triangles.
💡Takeaway. Find the whole length, find one triangle's length, then split into equal groups -- only Grade 4 grouping you already know!
Variant 3 easy answer: 5 triangles

A piece of wire is as long as a 5 cm5\ \text{cm} stick laid end to end 66 times. Using this wire, what is the greatest number of triangles you can make if each triangle has three sides that are all 2 cm2\ \text{cm} long?

Show solution
1 · Understandwhat's really being asked

A wire is as long as a 5 cm stick placed end to end 6 times. Each triangle is made of three 2 cm sides. We want the greatest number of such triangles the wire can make.

Givens
  • The wire length equals a 5 cm stick laid end to end 6 times.
  • Each triangle has three sides, each 2 cm long.
Unknowns
  • The greatest number of 2 cm equilateral triangles the wire can make.
Constraints
  • Each triangle uses exactly 3 sides of 2 cm.
  • Whole triangles only (you cannot use part of a triangle).
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units

Break the problem into small steps: first find the total wire length, then find how much wire one triangle needs, and finally see how many triangles fit. Tracking centimeters keeps the multiplication and division lined up correctly.

3 · Execute3 carry out the plan

1Find the total wire length

#7 Identify Subproblems 3.OA.B.5
The wire is a 5 cm length repeated 6 times, so multiply 5 by 6.
5×6=305 \times 6 = 30
6 equal lengths of 5 cm is just the multiplication fact 5 times 6.

2Find the wire one triangle needs

#8 Analyze the Units 3.OA.B.5
Each triangle has three 2 cm sides, so one triangle uses 2 cm three times.
2×3=62 \times 3 = 6
Three sides of the same length is simply 2 added three times, an easy times-three fact.

3Count how many triangles fit

#7 Identify Subproblems 4.OA.B.4
See how many 6 cm pieces fit in 30 cm. Because 6 times 5 equals 30, exactly 5 triangles can be made with no wire left over.
30÷6=530 \div 6 = 5
Finding how many equal groups of 6 make 30 is the same as knowing the factor pair 6 and 5 of 30.
Answer: 5 triangles
4 · Reviewdoes it hold up?

5 triangles use 6 times 5 = 30 cm, which is exactly the whole wire, so the answer fits with nothing wasted and the magnitude is sensible.

Another way: You could repeatedly subtract 6 cm from 30 cm and count the 5 subtractions, which is the repeated-subtraction view of the same division.

Standardsmin grade 4
  • 3.OA.B.5 Apply properties of operations as strategies to multiply and divide — Multiplying 5 by 6 for the total wire length and 2 by 3 for one triangle's wire.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that 30 splits into equal groups of 6, giving 5 triangles.
💡Takeaway. Find the whole length, find one triangle's length, then split into equal groups -- only Grade 4 grouping you already know!
Variant 4 easy answer: 7 triangles

A piece of wire is as long as a 7 cm7\ \text{cm} stick laid end to end 66 times. Using this wire, what is the greatest number of triangles you can make if each triangle has three sides that are all 2 cm2\ \text{cm} long?

Show solution
1 · Understandwhat's really being asked

A wire is as long as a 7 cm stick placed end to end 6 times. Each triangle is made of three 2 cm sides. We want the greatest number of such triangles the wire can make.

Givens
  • The wire length equals a 7 cm stick laid end to end 6 times.
  • Each triangle has three sides, each 2 cm long.
Unknowns
  • The greatest number of 2 cm equilateral triangles the wire can make.
Constraints
  • Each triangle uses exactly 3 sides of 2 cm.
  • Whole triangles only (you cannot use part of a triangle).
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units

Break the problem into small steps: first find the total wire length, then find how much wire one triangle needs, and finally see how many triangles fit. Tracking centimeters keeps the multiplication and division lined up correctly.

3 · Execute3 carry out the plan

1Find the total wire length

#7 Identify Subproblems 3.OA.B.5
The wire is a 7 cm length repeated 6 times, so multiply 7 by 6.
7×6=427 \times 6 = 42
6 equal lengths of 7 cm is just the multiplication fact 7 times 6.

2Find the wire one triangle needs

#8 Analyze the Units 3.OA.B.5
Each triangle has three 2 cm sides, so one triangle uses 2 cm three times.
2×3=62 \times 3 = 6
Three sides of the same length is simply 2 added three times, an easy times-three fact.

3Count how many triangles fit

#7 Identify Subproblems 4.OA.B.4
See how many 6 cm pieces fit in 42 cm. Because 6 times 7 equals 42, exactly 7 triangles can be made with no wire left over.
42÷6=742 \div 6 = 7
Finding how many equal groups of 6 make 42 is the same as knowing the factor pair 6 and 7 of 42.
Answer: 7 triangles
4 · Reviewdoes it hold up?

7 triangles use 6 times 7 = 42 cm, which is exactly the whole wire, so the answer fits with nothing wasted and the magnitude is sensible.

Another way: You could repeatedly subtract 6 cm from 42 cm and count the 7 subtractions, which is the repeated-subtraction view of the same division.

Standardsmin grade 4
  • 3.OA.B.5 Apply properties of operations as strategies to multiply and divide — Multiplying 7 by 6 for the total wire length and 2 by 3 for one triangle's wire.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that 42 splits into equal groups of 6, giving 7 triangles.
💡Takeaway. Find the whole length, find one triangle's length, then split into equal groups -- only Grade 4 grouping you already know!
Variant 5 medium answer: 4 triangles

A piece of wire is as long as a 8 cm8\ \text{cm} stick laid end to end 33 times. Using this wire, what is the greatest number of triangles you can make if each triangle has three sides that are all 2 cm2\ \text{cm} long?

Show solution
1 · Understandwhat's really being asked

A wire is as long as a 8 cm stick placed end to end 3 times. Each triangle is made of three 2 cm sides. We want the greatest number of such triangles the wire can make.

Givens
  • The wire length equals a 8 cm stick laid end to end 3 times.
  • Each triangle has three sides, each 2 cm long.
Unknowns
  • The greatest number of 2 cm equilateral triangles the wire can make.
Constraints
  • Each triangle uses exactly 3 sides of 2 cm.
  • Whole triangles only (you cannot use part of a triangle).
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units

Break the problem into small steps: first find the total wire length, then find how much wire one triangle needs, and finally see how many triangles fit. Tracking centimeters keeps the multiplication and division lined up correctly.

3 · Execute3 carry out the plan

1Find the total wire length

#7 Identify Subproblems 3.OA.B.5
The wire is a 8 cm length repeated 3 times, so multiply 8 by 3.
8×3=248 \times 3 = 24
3 equal lengths of 8 cm is just the multiplication fact 8 times 3.

2Find the wire one triangle needs

#8 Analyze the Units 3.OA.B.5
Each triangle has three 2 cm sides, so one triangle uses 2 cm three times.
2×3=62 \times 3 = 6
Three sides of the same length is simply 2 added three times, an easy times-three fact.

3Count how many triangles fit

#7 Identify Subproblems 4.OA.B.4
See how many 6 cm pieces fit in 24 cm. Because 6 times 4 equals 24, exactly 4 triangles can be made with no wire left over.
24÷6=424 \div 6 = 4
Finding how many equal groups of 6 make 24 is the same as knowing the factor pair 6 and 4 of 24.
Answer: 4 triangles
4 · Reviewdoes it hold up?

4 triangles use 6 times 4 = 24 cm, which is exactly the whole wire, so the answer fits with nothing wasted and the magnitude is sensible.

Another way: You could repeatedly subtract 6 cm from 24 cm and count the 4 subtractions, which is the repeated-subtraction view of the same division.

Standardsmin grade 4
  • 3.OA.B.5 Apply properties of operations as strategies to multiply and divide — Multiplying 8 by 3 for the total wire length and 2 by 3 for one triangle's wire.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that 24 splits into equal groups of 6, giving 4 triangles.
💡Takeaway. Find the whole length, find one triangle's length, then split into equal groups -- only Grade 4 grouping you already know!
Variant 6 medium answer: 4 triangles

A piece of wire is as long as a 8 cm8\ \text{cm} stick laid end to end 66 times. Using this wire, what is the greatest number of triangles you can make if each triangle has three sides that are all 4 cm4\ \text{cm} long?

Show solution
1 · Understandwhat's really being asked

A wire is as long as a 8 cm stick placed end to end 6 times. Each triangle is made of three 4 cm sides. We want the greatest number of such triangles the wire can make.

Givens
  • The wire length equals a 8 cm stick laid end to end 6 times.
  • Each triangle has three sides, each 4 cm long.
Unknowns
  • The greatest number of 4 cm equilateral triangles the wire can make.
Constraints
  • Each triangle uses exactly 3 sides of 4 cm.
  • Whole triangles only (you cannot use part of a triangle).
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units

Break the problem into small steps: first find the total wire length, then find how much wire one triangle needs, and finally see how many triangles fit. Tracking centimeters keeps the multiplication and division lined up correctly.

3 · Execute3 carry out the plan

1Find the total wire length

#7 Identify Subproblems 3.OA.B.5
The wire is a 8 cm length repeated 6 times, so multiply 8 by 6.
8×6=488 \times 6 = 48
6 equal lengths of 8 cm is just the multiplication fact 8 times 6.

2Find the wire one triangle needs

#8 Analyze the Units 3.OA.B.5
Each triangle has three 4 cm sides, so one triangle uses 4 cm three times.
4×3=124 \times 3 = 12
Three sides of the same length is simply 4 added three times, an easy times-three fact.

3Count how many triangles fit

#7 Identify Subproblems 4.OA.B.4
See how many 12 cm pieces fit in 48 cm. Because 12 times 4 equals 48, exactly 4 triangles can be made with no wire left over.
48÷12=448 \div 12 = 4
Finding how many equal groups of 12 make 48 is the same as knowing the factor pair 12 and 4 of 48.
Answer: 4 triangles
4 · Reviewdoes it hold up?

4 triangles use 12 times 4 = 48 cm, which is exactly the whole wire, so the answer fits with nothing wasted and the magnitude is sensible.

Another way: You could repeatedly subtract 12 cm from 48 cm and count the 4 subtractions, which is the repeated-subtraction view of the same division.

Standardsmin grade 4
  • 3.OA.B.5 Apply properties of operations as strategies to multiply and divide — Multiplying 8 by 6 for the total wire length and 4 by 3 for one triangle's wire.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that 48 splits into equal groups of 12, giving 4 triangles.
💡Takeaway. Find the whole length, find one triangle's length, then split into equal groups -- only Grade 4 grouping you already know!
Variant 7 medium answer: 2 triangles

A piece of wire is as long as a 9 cm9\ \text{cm} stick laid end to end 22 times. Using this wire, what is the greatest number of triangles you can make if each triangle has three sides that are all 3 cm3\ \text{cm} long?

Show solution
1 · Understandwhat's really being asked

A wire is as long as a 9 cm stick placed end to end 2 times. Each triangle is made of three 3 cm sides. We want the greatest number of such triangles the wire can make.

Givens
  • The wire length equals a 9 cm stick laid end to end 2 times.
  • Each triangle has three sides, each 3 cm long.
Unknowns
  • The greatest number of 3 cm equilateral triangles the wire can make.
Constraints
  • Each triangle uses exactly 3 sides of 3 cm.
  • Whole triangles only (you cannot use part of a triangle).
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units

Break the problem into small steps: first find the total wire length, then find how much wire one triangle needs, and finally see how many triangles fit. Tracking centimeters keeps the multiplication and division lined up correctly.

3 · Execute3 carry out the plan

1Find the total wire length

#7 Identify Subproblems 3.OA.B.5
The wire is a 9 cm length repeated 2 times, so multiply 9 by 2.
9×2=189 \times 2 = 18
2 equal lengths of 9 cm is just the multiplication fact 9 times 2.

2Find the wire one triangle needs

#8 Analyze the Units 3.OA.B.5
Each triangle has three 3 cm sides, so one triangle uses 3 cm three times.
3×3=93 \times 3 = 9
Three sides of the same length is simply 3 added three times, an easy times-three fact.

3Count how many triangles fit

#7 Identify Subproblems 4.OA.B.4
See how many 9 cm pieces fit in 18 cm. Because 9 times 2 equals 18, exactly 2 triangles can be made with no wire left over.
18÷9=218 \div 9 = 2
Finding how many equal groups of 9 make 18 is the same as knowing the factor pair 9 and 2 of 18.
Answer: 2 triangles
4 · Reviewdoes it hold up?

2 triangles use 9 times 2 = 18 cm, which is exactly the whole wire, so the answer fits with nothing wasted and the magnitude is sensible.

Another way: You could repeatedly subtract 9 cm from 18 cm and count the 2 subtractions, which is the repeated-subtraction view of the same division.

Standardsmin grade 4
  • 3.OA.B.5 Apply properties of operations as strategies to multiply and divide — Multiplying 9 by 2 for the total wire length and 3 by 3 for one triangle's wire.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that 18 splits into equal groups of 9, giving 2 triangles.
💡Takeaway. Find the whole length, find one triangle's length, then split into equal groups -- only Grade 4 grouping you already know!
Variant 8 medium answer: 6 triangles

A piece of wire is as long as a 9 cm9\ \text{cm} stick laid end to end 44 times. Using this wire, what is the greatest number of triangles you can make if each triangle has three sides that are all 2 cm2\ \text{cm} long?

Show solution
1 · Understandwhat's really being asked

A wire is as long as a 9 cm stick placed end to end 4 times. Each triangle is made of three 2 cm sides. We want the greatest number of such triangles the wire can make.

Givens
  • The wire length equals a 9 cm stick laid end to end 4 times.
  • Each triangle has three sides, each 2 cm long.
Unknowns
  • The greatest number of 2 cm equilateral triangles the wire can make.
Constraints
  • Each triangle uses exactly 3 sides of 2 cm.
  • Whole triangles only (you cannot use part of a triangle).
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units

Break the problem into small steps: first find the total wire length, then find how much wire one triangle needs, and finally see how many triangles fit. Tracking centimeters keeps the multiplication and division lined up correctly.

3 · Execute3 carry out the plan

1Find the total wire length

#7 Identify Subproblems 3.OA.B.5
The wire is a 9 cm length repeated 4 times, so multiply 9 by 4.
9×4=369 \times 4 = 36
4 equal lengths of 9 cm is just the multiplication fact 9 times 4.

2Find the wire one triangle needs

#8 Analyze the Units 3.OA.B.5
Each triangle has three 2 cm sides, so one triangle uses 2 cm three times.
2×3=62 \times 3 = 6
Three sides of the same length is simply 2 added three times, an easy times-three fact.

3Count how many triangles fit

#7 Identify Subproblems 4.OA.B.4
See how many 6 cm pieces fit in 36 cm. Because 6 times 6 equals 36, exactly 6 triangles can be made with no wire left over.
36÷6=636 \div 6 = 6
Finding how many equal groups of 6 make 36 is the same as knowing the factor pair 6 and 6 of 36.
Answer: 6 triangles
4 · Reviewdoes it hold up?

6 triangles use 6 times 6 = 36 cm, which is exactly the whole wire, so the answer fits with nothing wasted and the magnitude is sensible.

Another way: You could repeatedly subtract 6 cm from 36 cm and count the 6 subtractions, which is the repeated-subtraction view of the same division.

Standardsmin grade 4
  • 3.OA.B.5 Apply properties of operations as strategies to multiply and divide — Multiplying 9 by 4 for the total wire length and 2 by 3 for one triangle's wire.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that 36 splits into equal groups of 6, giving 6 triangles.
💡Takeaway. Find the whole length, find one triangle's length, then split into equal groups -- only Grade 4 grouping you already know!
Variant 9 hard answer: 6 triangles

A piece of wire is as long as a 4 cm4\ \text{cm} stick laid end to end 99 times. Using this wire, what is the greatest number of triangles you can make if each triangle has three sides that are all 2 cm2\ \text{cm} long?

Show solution
1 · Understandwhat's really being asked

A wire is as long as a 4 cm stick placed end to end 9 times. Each triangle is made of three 2 cm sides. We want the greatest number of such triangles the wire can make.

Givens
  • The wire length equals a 4 cm stick laid end to end 9 times.
  • Each triangle has three sides, each 2 cm long.
Unknowns
  • The greatest number of 2 cm equilateral triangles the wire can make.
Constraints
  • Each triangle uses exactly 3 sides of 2 cm.
  • Whole triangles only (you cannot use part of a triangle).
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units

Break the problem into small steps: first find the total wire length, then find how much wire one triangle needs, and finally see how many triangles fit. Tracking centimeters keeps the multiplication and division lined up correctly.

3 · Execute3 carry out the plan

1Find the total wire length

#7 Identify Subproblems 3.OA.B.5
The wire is a 4 cm length repeated 9 times, so multiply 4 by 9.
4×9=364 \times 9 = 36
9 equal lengths of 4 cm is just the multiplication fact 4 times 9.

2Find the wire one triangle needs

#8 Analyze the Units 3.OA.B.5
Each triangle has three 2 cm sides, so one triangle uses 2 cm three times.
2×3=62 \times 3 = 6
Three sides of the same length is simply 2 added three times, an easy times-three fact.

3Count how many triangles fit

#7 Identify Subproblems 4.OA.B.4
See how many 6 cm pieces fit in 36 cm. Because 6 times 6 equals 36, exactly 6 triangles can be made with no wire left over.
36÷6=636 \div 6 = 6
Finding how many equal groups of 6 make 36 is the same as knowing the factor pair 6 and 6 of 36.
Answer: 6 triangles
4 · Reviewdoes it hold up?

6 triangles use 6 times 6 = 36 cm, which is exactly the whole wire, so the answer fits with nothing wasted and the magnitude is sensible.

Another way: You could repeatedly subtract 6 cm from 36 cm and count the 6 subtractions, which is the repeated-subtraction view of the same division.

Standardsmin grade 4
  • 3.OA.B.5 Apply properties of operations as strategies to multiply and divide — Multiplying 4 by 9 for the total wire length and 2 by 3 for one triangle's wire.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that 36 splits into equal groups of 6, giving 6 triangles.
💡Takeaway. Find the whole length, find one triangle's length, then split into equal groups -- only Grade 4 grouping you already know!
Variant 10 hard answer: 8 triangles

A piece of wire is as long as a 9 cm9\ \text{cm} stick laid end to end 88 times. Using this wire, what is the greatest number of triangles you can make if each triangle has three sides that are all 3 cm3\ \text{cm} long?

Show solution
1 · Understandwhat's really being asked

A wire is as long as a 9 cm stick placed end to end 8 times. Each triangle is made of three 3 cm sides. We want the greatest number of such triangles the wire can make.

Givens
  • The wire length equals a 9 cm stick laid end to end 8 times.
  • Each triangle has three sides, each 3 cm long.
Unknowns
  • The greatest number of 3 cm equilateral triangles the wire can make.
Constraints
  • Each triangle uses exactly 3 sides of 3 cm.
  • Whole triangles only (you cannot use part of a triangle).
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units

Break the problem into small steps: first find the total wire length, then find how much wire one triangle needs, and finally see how many triangles fit. Tracking centimeters keeps the multiplication and division lined up correctly.

3 · Execute3 carry out the plan

1Find the total wire length

#7 Identify Subproblems 3.OA.B.5
The wire is a 9 cm length repeated 8 times, so multiply 9 by 8.
9×8=729 \times 8 = 72
8 equal lengths of 9 cm is just the multiplication fact 9 times 8.

2Find the wire one triangle needs

#8 Analyze the Units 3.OA.B.5
Each triangle has three 3 cm sides, so one triangle uses 3 cm three times.
3×3=93 \times 3 = 9
Three sides of the same length is simply 3 added three times, an easy times-three fact.

3Count how many triangles fit

#7 Identify Subproblems 4.OA.B.4
See how many 9 cm pieces fit in 72 cm. Because 9 times 8 equals 72, exactly 8 triangles can be made with no wire left over.
72÷9=872 \div 9 = 8
Finding how many equal groups of 9 make 72 is the same as knowing the factor pair 9 and 8 of 72.
Answer: 8 triangles
4 · Reviewdoes it hold up?

8 triangles use 9 times 8 = 72 cm, which is exactly the whole wire, so the answer fits with nothing wasted and the magnitude is sensible.

Another way: You could repeatedly subtract 9 cm from 72 cm and count the 8 subtractions, which is the repeated-subtraction view of the same division.

Standardsmin grade 4
  • 3.OA.B.5 Apply properties of operations as strategies to multiply and divide — Multiplying 9 by 8 for the total wire length and 3 by 3 for one triangle's wire.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that 72 splits into equal groups of 9, giving 8 triangles.
💡Takeaway. Find the whole length, find one triangle's length, then split into equal groups -- only Grade 4 grouping you already know!
Variant 11 hard answer: 5 triangles

A piece of wire is as long as a 10 cm10\ \text{cm} stick laid end to end 33 times. Using this wire, what is the greatest number of triangles you can make if each triangle has three sides that are all 2 cm2\ \text{cm} long?

Show solution
1 · Understandwhat's really being asked

A wire is as long as a 10 cm stick placed end to end 3 times. Each triangle is made of three 2 cm sides. We want the greatest number of such triangles the wire can make.

Givens
  • The wire length equals a 10 cm stick laid end to end 3 times.
  • Each triangle has three sides, each 2 cm long.
Unknowns
  • The greatest number of 2 cm equilateral triangles the wire can make.
Constraints
  • Each triangle uses exactly 3 sides of 2 cm.
  • Whole triangles only (you cannot use part of a triangle).
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units

Break the problem into small steps: first find the total wire length, then find how much wire one triangle needs, and finally see how many triangles fit. Tracking centimeters keeps the multiplication and division lined up correctly.

3 · Execute3 carry out the plan

1Find the total wire length

#7 Identify Subproblems 3.OA.B.5
The wire is a 10 cm length repeated 3 times, so multiply 10 by 3.
10×3=3010 \times 3 = 30
3 equal lengths of 10 cm is just the multiplication fact 10 times 3.

2Find the wire one triangle needs

#8 Analyze the Units 3.OA.B.5
Each triangle has three 2 cm sides, so one triangle uses 2 cm three times.
2×3=62 \times 3 = 6
Three sides of the same length is simply 2 added three times, an easy times-three fact.

3Count how many triangles fit

#7 Identify Subproblems 4.OA.B.4
See how many 6 cm pieces fit in 30 cm. Because 6 times 5 equals 30, exactly 5 triangles can be made with no wire left over.
30÷6=530 \div 6 = 5
Finding how many equal groups of 6 make 30 is the same as knowing the factor pair 6 and 5 of 30.
Answer: 5 triangles
4 · Reviewdoes it hold up?

5 triangles use 6 times 5 = 30 cm, which is exactly the whole wire, so the answer fits with nothing wasted and the magnitude is sensible.

Another way: You could repeatedly subtract 6 cm from 30 cm and count the 5 subtractions, which is the repeated-subtraction view of the same division.

Standardsmin grade 4
  • 3.OA.B.5 Apply properties of operations as strategies to multiply and divide — Multiplying 10 by 3 for the total wire length and 2 by 3 for one triangle's wire.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that 30 splits into equal groups of 6, giving 5 triangles.
💡Takeaway. Find the whole length, find one triangle's length, then split into equal groups -- only Grade 4 grouping you already know!
Variant 12 hard answer: 4 triangles

A piece of wire is as long as a 12 cm12\ \text{cm} stick laid end to end 44 times. Using this wire, what is the greatest number of triangles you can make if each triangle has three sides that are all 4 cm4\ \text{cm} long?

Show solution
1 · Understandwhat's really being asked

A wire is as long as a 12 cm stick placed end to end 4 times. Each triangle is made of three 4 cm sides. We want the greatest number of such triangles the wire can make.

Givens
  • The wire length equals a 12 cm stick laid end to end 4 times.
  • Each triangle has three sides, each 4 cm long.
Unknowns
  • The greatest number of 4 cm equilateral triangles the wire can make.
Constraints
  • Each triangle uses exactly 3 sides of 4 cm.
  • Whole triangles only (you cannot use part of a triangle).
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units

Break the problem into small steps: first find the total wire length, then find how much wire one triangle needs, and finally see how many triangles fit. Tracking centimeters keeps the multiplication and division lined up correctly.

3 · Execute3 carry out the plan

1Find the total wire length

#7 Identify Subproblems 3.OA.B.5
The wire is a 12 cm length repeated 4 times, so multiply 12 by 4.
12×4=4812 \times 4 = 48
4 equal lengths of 12 cm is just the multiplication fact 12 times 4.

2Find the wire one triangle needs

#8 Analyze the Units 3.OA.B.5
Each triangle has three 4 cm sides, so one triangle uses 4 cm three times.
4×3=124 \times 3 = 12
Three sides of the same length is simply 4 added three times, an easy times-three fact.

3Count how many triangles fit

#7 Identify Subproblems 4.OA.B.4
See how many 12 cm pieces fit in 48 cm. Because 12 times 4 equals 48, exactly 4 triangles can be made with no wire left over.
48÷12=448 \div 12 = 4
Finding how many equal groups of 12 make 48 is the same as knowing the factor pair 12 and 4 of 48.
Answer: 4 triangles
4 · Reviewdoes it hold up?

4 triangles use 12 times 4 = 48 cm, which is exactly the whole wire, so the answer fits with nothing wasted and the magnitude is sensible.

Another way: You could repeatedly subtract 12 cm from 48 cm and count the 4 subtractions, which is the repeated-subtraction view of the same division.

Standardsmin grade 4
  • 3.OA.B.5 Apply properties of operations as strategies to multiply and divide — Multiplying 12 by 4 for the total wire length and 4 by 3 for one triangle's wire.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Recognizing that 48 splits into equal groups of 12, giving 4 triangles.
💡Takeaway. Find the whole length, find one triangle's length, then split into equal groups -- only Grade 4 grouping you already know!