← Meshing gears realign at LCM of teeth · Repeating Cycle Patterns

Meshing gears realign at LCM of teeth · 12 practice problems

6.NS.B.44.OA.B.44.OA.C.5

Generated variants — 12

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

Variant 1 easy answer: Gear A: 3 turns, Gear B: 4 turns

Two gears are meshed together: gear A has 2020 teeth and gear B has 1515 teeth. The gears begin turning together from the point where they are first meshed. To return to the position where the same two teeth mesh again, at least how many full turns must gear A and gear B each make?

Show solution
1 · Understandwhat's really being asked

Gear A has 20 teeth and gear B has 15. They start with one particular pair of teeth touching. We need how many whole turns each makes before that same pair touches again.

Givens
  • Gear A has 20 teeth.
  • Gear B has 15 teeth.
  • They start meshed at a marked pair of teeth.
Unknowns
  • The number of full turns each gear makes before the marked pair meets again.
Constraints
  • Meshed gears pass teeth one for one, so both gears count the same number of teeth through the mesh point.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Count teeth, not turns. Each gear is back where it started after a multiple of its own tooth count, so the marked pair meets again at the first count that is a multiple of both.

3 · Execute3 carry out the plan

1See when each gear comes home

#9 Solve an Easier Related Problem 4.OA.B.4
Gear A is back at its starting position after every 20 teeth pass, gear B after every 15.
A:20,40,60, ⋯B:15,30,45,60, ⋯A: 20, 40, 60,\ \cdots\quad B: 15, 30, 45, 60,\ \cdots
Two lists of multiples, one per gear.

2Find the first shared count (the LCM)

#5 Look for a Pattern 6.NS.B.4
Both gears are home together at the first number on both lists.
lcm⁡(20,15)=60\operatorname{lcm}(20, 15) = 60
After 60 teeth the marked pair meets again.

3Turn the 60 teeth into full turns for each gear

#2 Make a Systematic List 4.OA.C.5
Divide the shared tooth count by each gear's own tooth count.
60÷20=3,60÷15=460 \div 20 = 3,\quad 60 \div 15 = 4
Gear A turns 3 times, gear B 4 times.
Answer: Gear A: 3 turns, Gear B: 4 turns
4 · Reviewdoes it hold up?

Both come to the same tooth count: 20 x 3 = 60 and 15 x 4 = 60. Gear B is the smaller one and turns the most, 4 times, which is what a smaller gear does.

Another way: Divide both tooth counts by their greatest common factor, 5: that gives 4 and 3, and the turns are those two numbers swapped.

Standardsmin grade 6
  • 6.NS.B.4 Find greatest common factor and least common multiple of two numbers — Finding 60 as the first tooth count both gears share.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing each gear's multiples of teeth.
  • 4.OA.C.5 Generate a number or shape pattern following a given rule — Converting the shared tooth count into turns for each gear.
💡Takeaway. Meshed gears count the same teeth, not the same turns -- so the smaller gear has to spin more to get back home.
Variant 2 easy answer: Gear A: 3 turns, Gear B: 4 turns

Two gears are meshed together: gear A has 2424 teeth and gear B has 1818 teeth. The gears begin turning together from the point where they are first meshed. To return to the position where the same two teeth mesh again, at least how many full turns must gear A and gear B each make?

Show solution
1 · Understandwhat's really being asked

Gear A has 24 teeth and gear B has 18. They start with one particular pair of teeth touching. We need how many whole turns each makes before that same pair touches again.

Givens
  • Gear A has 24 teeth.
  • Gear B has 18 teeth.
  • They start meshed at a marked pair of teeth.
Unknowns
  • The number of full turns each gear makes before the marked pair meets again.
Constraints
  • Meshed gears pass teeth one for one, so both gears count the same number of teeth through the mesh point.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Count teeth, not turns. Each gear is back where it started after a multiple of its own tooth count, so the marked pair meets again at the first count that is a multiple of both.

3 · Execute3 carry out the plan

1See when each gear comes home

#9 Solve an Easier Related Problem 4.OA.B.4
Gear A is back at its starting position after every 24 teeth pass, gear B after every 18.
A:24,48,72, ⋯B:18,36,54,72, ⋯A: 24, 48, 72,\ \cdots\quad B: 18, 36, 54, 72,\ \cdots
Two lists of multiples, one per gear.

2Find the first shared count (the LCM)

#5 Look for a Pattern 6.NS.B.4
Both gears are home together at the first number on both lists.
lcm⁡(24,18)=72\operatorname{lcm}(24, 18) = 72
After 72 teeth the marked pair meets again.

3Turn the 72 teeth into full turns for each gear

#2 Make a Systematic List 4.OA.C.5
Divide the shared tooth count by each gear's own tooth count.
72÷24=3,72÷18=472 \div 24 = 3,\quad 72 \div 18 = 4
Gear A turns 3 times, gear B 4 times.
Answer: Gear A: 3 turns, Gear B: 4 turns
4 · Reviewdoes it hold up?

Both come to the same tooth count: 24 x 3 = 72 and 18 x 4 = 72. Gear B is the smaller one and turns the most, 4 times, which is what a smaller gear does.

Another way: Divide both tooth counts by their greatest common factor, 6: that gives 4 and 3, and the turns are those two numbers swapped.

Standardsmin grade 6
  • 6.NS.B.4 Find greatest common factor and least common multiple of two numbers — Finding 72 as the first tooth count both gears share.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing each gear's multiples of teeth.
  • 4.OA.C.5 Generate a number or shape pattern following a given rule — Converting the shared tooth count into turns for each gear.
💡Takeaway. Meshed gears count the same teeth, not the same turns -- so the smaller gear has to spin more to get back home.
Variant 3 easy answer: Gear A: 3 turns, Gear B: 4 turns

Two gears are meshed together: gear A has 2828 teeth and gear B has 2121 teeth. The gears begin turning together from the point where they are first meshed. To return to the position where the same two teeth mesh again, at least how many full turns must gear A and gear B each make?

Show solution
1 · Understandwhat's really being asked

Gear A has 28 teeth and gear B has 21. They start with one particular pair of teeth touching. We need how many whole turns each makes before that same pair touches again.

Givens
  • Gear A has 28 teeth.
  • Gear B has 21 teeth.
  • They start meshed at a marked pair of teeth.
Unknowns
  • The number of full turns each gear makes before the marked pair meets again.
Constraints
  • Meshed gears pass teeth one for one, so both gears count the same number of teeth through the mesh point.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Count teeth, not turns. Each gear is back where it started after a multiple of its own tooth count, so the marked pair meets again at the first count that is a multiple of both.

3 · Execute3 carry out the plan

1See when each gear comes home

#9 Solve an Easier Related Problem 4.OA.B.4
Gear A is back at its starting position after every 28 teeth pass, gear B after every 21.
A:28,56,84, ⋯B:21,42,63,84, ⋯A: 28, 56, 84,\ \cdots\quad B: 21, 42, 63, 84,\ \cdots
Two lists of multiples, one per gear.

2Find the first shared count (the LCM)

#5 Look for a Pattern 6.NS.B.4
Both gears are home together at the first number on both lists.
lcm⁡(28,21)=84\operatorname{lcm}(28, 21) = 84
After 84 teeth the marked pair meets again.

3Turn the 84 teeth into full turns for each gear

#2 Make a Systematic List 4.OA.C.5
Divide the shared tooth count by each gear's own tooth count.
84÷28=3,84÷21=484 \div 28 = 3,\quad 84 \div 21 = 4
Gear A turns 3 times, gear B 4 times.
Answer: Gear A: 3 turns, Gear B: 4 turns
4 · Reviewdoes it hold up?

Both come to the same tooth count: 28 x 3 = 84 and 21 x 4 = 84. Gear B is the smaller one and turns the most, 4 times, which is what a smaller gear does.

Another way: Divide both tooth counts by their greatest common factor, 7: that gives 4 and 3, and the turns are those two numbers swapped.

Standardsmin grade 6
  • 6.NS.B.4 Find greatest common factor and least common multiple of two numbers — Finding 84 as the first tooth count both gears share.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing each gear's multiples of teeth.
  • 4.OA.C.5 Generate a number or shape pattern following a given rule — Converting the shared tooth count into turns for each gear.
💡Takeaway. Meshed gears count the same teeth, not the same turns -- so the smaller gear has to spin more to get back home.
Variant 4 easy answer: Gear A: 3 turns, Gear B: 5 turns

Two gears are meshed together: gear A has 3030 teeth and gear B has 1818 teeth. The gears begin turning together from the point where they are first meshed. To return to the position where the same two teeth mesh again, at least how many full turns must gear A and gear B each make?

Show solution
1 · Understandwhat's really being asked

Gear A has 30 teeth and gear B has 18. They start with one particular pair of teeth touching. We need how many whole turns each makes before that same pair touches again.

Givens
  • Gear A has 30 teeth.
  • Gear B has 18 teeth.
  • They start meshed at a marked pair of teeth.
Unknowns
  • The number of full turns each gear makes before the marked pair meets again.
Constraints
  • Meshed gears pass teeth one for one, so both gears count the same number of teeth through the mesh point.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Count teeth, not turns. Each gear is back where it started after a multiple of its own tooth count, so the marked pair meets again at the first count that is a multiple of both.

3 · Execute3 carry out the plan

1See when each gear comes home

#9 Solve an Easier Related Problem 4.OA.B.4
Gear A is back at its starting position after every 30 teeth pass, gear B after every 18.
A:30,60,90, ⋯B:18,36,54,72,90, ⋯A: 30, 60, 90,\ \cdots\quad B: 18, 36, 54, 72, 90,\ \cdots
Two lists of multiples, one per gear.

2Find the first shared count (the LCM)

#5 Look for a Pattern 6.NS.B.4
Both gears are home together at the first number on both lists.
lcm⁡(30,18)=90\operatorname{lcm}(30, 18) = 90
After 90 teeth the marked pair meets again.

3Turn the 90 teeth into full turns for each gear

#2 Make a Systematic List 4.OA.C.5
Divide the shared tooth count by each gear's own tooth count.
90÷30=3,90÷18=590 \div 30 = 3,\quad 90 \div 18 = 5
Gear A turns 3 times, gear B 5 times.
Answer: Gear A: 3 turns, Gear B: 5 turns
4 · Reviewdoes it hold up?

Both come to the same tooth count: 30 x 3 = 90 and 18 x 5 = 90. Gear B is the smaller one and turns the most, 5 times, which is what a smaller gear does.

Another way: Divide both tooth counts by their greatest common factor, 6: that gives 5 and 3, and the turns are those two numbers swapped.

Standardsmin grade 6
  • 6.NS.B.4 Find greatest common factor and least common multiple of two numbers — Finding 90 as the first tooth count both gears share.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing each gear's multiples of teeth.
  • 4.OA.C.5 Generate a number or shape pattern following a given rule — Converting the shared tooth count into turns for each gear.
💡Takeaway. Meshed gears count the same teeth, not the same turns -- so the smaller gear has to spin more to get back home.
Variant 5 medium answer: Gear A: 5 turns, Gear B: 8 turns

Two gears are meshed together: gear A has 3232 teeth and gear B has 2020 teeth. The gears begin turning together from the point where they are first meshed. To return to the position where the same two teeth mesh again, at least how many full turns must gear A and gear B each make?

Show solution
1 · Understandwhat's really being asked

Gear A has 32 teeth and gear B has 20. They start with one particular pair of teeth touching. We need how many whole turns each makes before that same pair touches again.

Givens
  • Gear A has 32 teeth.
  • Gear B has 20 teeth.
  • They start meshed at a marked pair of teeth.
Unknowns
  • The number of full turns each gear makes before the marked pair meets again.
Constraints
  • Meshed gears pass teeth one for one, so both gears count the same number of teeth through the mesh point.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Count teeth, not turns. Each gear is back where it started after a multiple of its own tooth count, so the marked pair meets again at the first count that is a multiple of both.

3 · Execute3 carry out the plan

1See when each gear comes home

#9 Solve an Easier Related Problem 4.OA.B.4
Gear A is back at its starting position after every 32 teeth pass, gear B after every 20.
A:32,64,96,128,160, ⋯B:20,40,60,80,100,120,140,160, ⋯A: 32, 64, 96, 128, 160,\ \cdots\quad B: 20, 40, 60, 80, 100, 120, 140, 160,\ \cdots
Two lists of multiples, one per gear.

2Find the first shared count (the LCM)

#5 Look for a Pattern 6.NS.B.4
Both gears are home together at the first number on both lists.
lcm⁡(32,20)=160\operatorname{lcm}(32, 20) = 160
After 160 teeth the marked pair meets again.

3Turn the 160 teeth into full turns for each gear

#2 Make a Systematic List 4.OA.C.5
Divide the shared tooth count by each gear's own tooth count.
160÷32=5,160÷20=8160 \div 32 = 5,\quad 160 \div 20 = 8
Gear A turns 5 times, gear B 8 times.
Answer: Gear A: 5 turns, Gear B: 8 turns
4 · Reviewdoes it hold up?

Both come to the same tooth count: 32 x 5 = 160 and 20 x 8 = 160. Gear B is the smaller one and turns the most, 8 times, which is what a smaller gear does.

Another way: Divide both tooth counts by their greatest common factor, 4: that gives 8 and 5, and the turns are those two numbers swapped.

Standardsmin grade 6
  • 6.NS.B.4 Find greatest common factor and least common multiple of two numbers — Finding 160 as the first tooth count both gears share.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing each gear's multiples of teeth.
  • 4.OA.C.5 Generate a number or shape pattern following a given rule — Converting the shared tooth count into turns for each gear.
💡Takeaway. Meshed gears count the same teeth, not the same turns -- so the smaller gear has to spin more to get back home.
Variant 6 medium answer: Gear A: 5 turns, Gear B: 7 turns

Two gears are meshed together: gear A has 3535 teeth and gear B has 2525 teeth. The gears begin turning together from the point where they are first meshed. To return to the position where the same two teeth mesh again, at least how many full turns must gear A and gear B each make?

Show solution
1 · Understandwhat's really being asked

Gear A has 35 teeth and gear B has 25. They start with one particular pair of teeth touching. We need how many whole turns each makes before that same pair touches again.

Givens
  • Gear A has 35 teeth.
  • Gear B has 25 teeth.
  • They start meshed at a marked pair of teeth.
Unknowns
  • The number of full turns each gear makes before the marked pair meets again.
Constraints
  • Meshed gears pass teeth one for one, so both gears count the same number of teeth through the mesh point.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Count teeth, not turns. Each gear is back where it started after a multiple of its own tooth count, so the marked pair meets again at the first count that is a multiple of both.

3 · Execute3 carry out the plan

1See when each gear comes home

#9 Solve an Easier Related Problem 4.OA.B.4
Gear A is back at its starting position after every 35 teeth pass, gear B after every 25.
A:35,70,105,140,175, ⋯B:25,50,75,100,125,150,175, ⋯A: 35, 70, 105, 140, 175,\ \cdots\quad B: 25, 50, 75, 100, 125, 150, 175,\ \cdots
Two lists of multiples, one per gear.

2Find the first shared count (the LCM)

#5 Look for a Pattern 6.NS.B.4
Both gears are home together at the first number on both lists.
lcm⁡(35,25)=175\operatorname{lcm}(35, 25) = 175
After 175 teeth the marked pair meets again.

3Turn the 175 teeth into full turns for each gear

#2 Make a Systematic List 4.OA.C.5
Divide the shared tooth count by each gear's own tooth count.
175÷35=5,175÷25=7175 \div 35 = 5,\quad 175 \div 25 = 7
Gear A turns 5 times, gear B 7 times.
Answer: Gear A: 5 turns, Gear B: 7 turns
4 · Reviewdoes it hold up?

Both come to the same tooth count: 35 x 5 = 175 and 25 x 7 = 175. Gear B is the smaller one and turns the most, 7 times, which is what a smaller gear does.

Another way: Divide both tooth counts by their greatest common factor, 5: that gives 7 and 5, and the turns are those two numbers swapped.

Standardsmin grade 6
  • 6.NS.B.4 Find greatest common factor and least common multiple of two numbers — Finding 175 as the first tooth count both gears share.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing each gear's multiples of teeth.
  • 4.OA.C.5 Generate a number or shape pattern following a given rule — Converting the shared tooth count into turns for each gear.
💡Takeaway. Meshed gears count the same teeth, not the same turns -- so the smaller gear has to spin more to get back home.
Variant 7 medium answer: Gear A: 2 turns, Gear B: 3 turns

Two gears are meshed together: gear A has 3636 teeth and gear B has 2424 teeth. The gears begin turning together from the point where they are first meshed. To return to the position where the same two teeth mesh again, at least how many full turns must gear A and gear B each make?

Show solution
1 · Understandwhat's really being asked

Gear A has 36 teeth and gear B has 24. They start with one particular pair of teeth touching. We need how many whole turns each makes before that same pair touches again.

Givens
  • Gear A has 36 teeth.
  • Gear B has 24 teeth.
  • They start meshed at a marked pair of teeth.
Unknowns
  • The number of full turns each gear makes before the marked pair meets again.
Constraints
  • Meshed gears pass teeth one for one, so both gears count the same number of teeth through the mesh point.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Count teeth, not turns. Each gear is back where it started after a multiple of its own tooth count, so the marked pair meets again at the first count that is a multiple of both.

3 · Execute3 carry out the plan

1See when each gear comes home

#9 Solve an Easier Related Problem 4.OA.B.4
Gear A is back at its starting position after every 36 teeth pass, gear B after every 24.
A:36,72, ⋯B:24,48,72, ⋯A: 36, 72,\ \cdots\quad B: 24, 48, 72,\ \cdots
Two lists of multiples, one per gear.

2Find the first shared count (the LCM)

#5 Look for a Pattern 6.NS.B.4
Both gears are home together at the first number on both lists.
lcm⁡(36,24)=72\operatorname{lcm}(36, 24) = 72
After 72 teeth the marked pair meets again.

3Turn the 72 teeth into full turns for each gear

#2 Make a Systematic List 4.OA.C.5
Divide the shared tooth count by each gear's own tooth count.
72÷36=2,72÷24=372 \div 36 = 2,\quad 72 \div 24 = 3
Gear A turns 2 times, gear B 3 times.
Answer: Gear A: 2 turns, Gear B: 3 turns
4 · Reviewdoes it hold up?

Both come to the same tooth count: 36 x 2 = 72 and 24 x 3 = 72. Gear B is the smaller one and turns the most, 3 times, which is what a smaller gear does.

Another way: Divide both tooth counts by their greatest common factor, 12: that gives 3 and 2, and the turns are those two numbers swapped.

Standardsmin grade 6
  • 6.NS.B.4 Find greatest common factor and least common multiple of two numbers — Finding 72 as the first tooth count both gears share.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing each gear's multiples of teeth.
  • 4.OA.C.5 Generate a number or shape pattern following a given rule — Converting the shared tooth count into turns for each gear.
💡Takeaway. Meshed gears count the same teeth, not the same turns -- so the smaller gear has to spin more to get back home.
Variant 8 medium answer: Gear A: 3 turns, Gear B: 2 turns

Two gears are meshed together: gear A has 2626 teeth and gear B has 3939 teeth. The gears begin turning together from the point where they are first meshed. To return to the position where the same two teeth mesh again, at least how many full turns must gear A and gear B each make?

Show solution
1 · Understandwhat's really being asked

Gear A has 26 teeth and gear B has 39. They start with one particular pair of teeth touching. We need how many whole turns each makes before that same pair touches again.

Givens
  • Gear A has 26 teeth.
  • Gear B has 39 teeth.
  • They start meshed at a marked pair of teeth.
Unknowns
  • The number of full turns each gear makes before the marked pair meets again.
Constraints
  • Meshed gears pass teeth one for one, so both gears count the same number of teeth through the mesh point.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Count teeth, not turns. Each gear is back where it started after a multiple of its own tooth count, so the marked pair meets again at the first count that is a multiple of both.

3 · Execute3 carry out the plan

1See when each gear comes home

#9 Solve an Easier Related Problem 4.OA.B.4
Gear A is back at its starting position after every 26 teeth pass, gear B after every 39.
A:26,52,78, ⋯B:39,78, ⋯A: 26, 52, 78,\ \cdots\quad B: 39, 78,\ \cdots
Two lists of multiples, one per gear.

2Find the first shared count (the LCM)

#5 Look for a Pattern 6.NS.B.4
Both gears are home together at the first number on both lists.
lcm⁡(26,39)=78\operatorname{lcm}(26, 39) = 78
After 78 teeth the marked pair meets again.

3Turn the 78 teeth into full turns for each gear

#2 Make a Systematic List 4.OA.C.5
Divide the shared tooth count by each gear's own tooth count.
78÷26=3,78÷39=278 \div 26 = 3,\quad 78 \div 39 = 2
Gear A turns 3 times, gear B 2 times.
Answer: Gear A: 3 turns, Gear B: 2 turns
4 · Reviewdoes it hold up?

Both come to the same tooth count: 26 x 3 = 78 and 39 x 2 = 78. Gear A is the smaller one and turns the most, 3 times, which is what a smaller gear does.

Another way: Divide both tooth counts by their greatest common factor, 13: that gives 2 and 3, and the turns are those two numbers swapped.

Standardsmin grade 6
  • 6.NS.B.4 Find greatest common factor and least common multiple of two numbers — Finding 78 as the first tooth count both gears share.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing each gear's multiples of teeth.
  • 4.OA.C.5 Generate a number or shape pattern following a given rule — Converting the shared tooth count into turns for each gear.
💡Takeaway. Meshed gears count the same teeth, not the same turns -- so the smaller gear has to spin more to get back home.
Variant 9 hard answer: Gear A: 3 turns, Gear B: 5 turns

Two gears are meshed together: gear A has 4040 teeth and gear B has 2424 teeth. The gears begin turning together from the point where they are first meshed. To return to the position where the same two teeth mesh again, at least how many full turns must gear A and gear B each make?

Show solution
1 · Understandwhat's really being asked

Gear A has 40 teeth and gear B has 24. They start with one particular pair of teeth touching. We need how many whole turns each makes before that same pair touches again.

Givens
  • Gear A has 40 teeth.
  • Gear B has 24 teeth.
  • They start meshed at a marked pair of teeth.
Unknowns
  • The number of full turns each gear makes before the marked pair meets again.
Constraints
  • Meshed gears pass teeth one for one, so both gears count the same number of teeth through the mesh point.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Count teeth, not turns. Each gear is back where it started after a multiple of its own tooth count, so the marked pair meets again at the first count that is a multiple of both.

3 · Execute3 carry out the plan

1See when each gear comes home

#9 Solve an Easier Related Problem 4.OA.B.4
Gear A is back at its starting position after every 40 teeth pass, gear B after every 24.
A:40,80,120, ⋯B:24,48,72,96,120, ⋯A: 40, 80, 120,\ \cdots\quad B: 24, 48, 72, 96, 120,\ \cdots
Two lists of multiples, one per gear.

2Find the first shared count (the LCM)

#5 Look for a Pattern 6.NS.B.4
Both gears are home together at the first number on both lists.
lcm⁡(40,24)=120\operatorname{lcm}(40, 24) = 120
After 120 teeth the marked pair meets again.

3Turn the 120 teeth into full turns for each gear

#2 Make a Systematic List 4.OA.C.5
Divide the shared tooth count by each gear's own tooth count.
120÷40=3,120÷24=5120 \div 40 = 3,\quad 120 \div 24 = 5
Gear A turns 3 times, gear B 5 times.
Answer: Gear A: 3 turns, Gear B: 5 turns
4 · Reviewdoes it hold up?

Both come to the same tooth count: 40 x 3 = 120 and 24 x 5 = 120. Gear B is the smaller one and turns the most, 5 times, which is what a smaller gear does.

Another way: Divide both tooth counts by their greatest common factor, 8: that gives 5 and 3, and the turns are those two numbers swapped.

Standardsmin grade 6
  • 6.NS.B.4 Find greatest common factor and least common multiple of two numbers — Finding 120 as the first tooth count both gears share.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing each gear's multiples of teeth.
  • 4.OA.C.5 Generate a number or shape pattern following a given rule — Converting the shared tooth count into turns for each gear.
💡Takeaway. Meshed gears count the same teeth, not the same turns -- so the smaller gear has to spin more to get back home.
Variant 10 hard answer: Gear A: 2 turns, Gear B: 3 turns

Two gears are meshed together: gear A has 4545 teeth and gear B has 3030 teeth. The gears begin turning together from the point where they are first meshed. To return to the position where the same two teeth mesh again, at least how many full turns must gear A and gear B each make?

Show solution
1 · Understandwhat's really being asked

Gear A has 45 teeth and gear B has 30. They start with one particular pair of teeth touching. We need how many whole turns each makes before that same pair touches again.

Givens
  • Gear A has 45 teeth.
  • Gear B has 30 teeth.
  • They start meshed at a marked pair of teeth.
Unknowns
  • The number of full turns each gear makes before the marked pair meets again.
Constraints
  • Meshed gears pass teeth one for one, so both gears count the same number of teeth through the mesh point.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Count teeth, not turns. Each gear is back where it started after a multiple of its own tooth count, so the marked pair meets again at the first count that is a multiple of both.

3 · Execute3 carry out the plan

1See when each gear comes home

#9 Solve an Easier Related Problem 4.OA.B.4
Gear A is back at its starting position after every 45 teeth pass, gear B after every 30.
A:45,90, ⋯B:30,60,90, ⋯A: 45, 90,\ \cdots\quad B: 30, 60, 90,\ \cdots
Two lists of multiples, one per gear.

2Find the first shared count (the LCM)

#5 Look for a Pattern 6.NS.B.4
Both gears are home together at the first number on both lists.
lcm⁡(45,30)=90\operatorname{lcm}(45, 30) = 90
After 90 teeth the marked pair meets again.

3Turn the 90 teeth into full turns for each gear

#2 Make a Systematic List 4.OA.C.5
Divide the shared tooth count by each gear's own tooth count.
90÷45=2,90÷30=390 \div 45 = 2,\quad 90 \div 30 = 3
Gear A turns 2 times, gear B 3 times.
Answer: Gear A: 2 turns, Gear B: 3 turns
4 · Reviewdoes it hold up?

Both come to the same tooth count: 45 x 2 = 90 and 30 x 3 = 90. Gear B is the smaller one and turns the most, 3 times, which is what a smaller gear does.

Another way: Divide both tooth counts by their greatest common factor, 15: that gives 3 and 2, and the turns are those two numbers swapped.

Standardsmin grade 6
  • 6.NS.B.4 Find greatest common factor and least common multiple of two numbers — Finding 90 as the first tooth count both gears share.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing each gear's multiples of teeth.
  • 4.OA.C.5 Generate a number or shape pattern following a given rule — Converting the shared tooth count into turns for each gear.
💡Takeaway. Meshed gears count the same teeth, not the same turns -- so the smaller gear has to spin more to get back home.
Variant 11 hard answer: Gear A: 2 turns, Gear B: 3 turns

Two gears are meshed together: gear A has 4848 teeth and gear B has 3232 teeth. The gears begin turning together from the point where they are first meshed. To return to the position where the same two teeth mesh again, at least how many full turns must gear A and gear B each make?

Show solution
1 · Understandwhat's really being asked

Gear A has 48 teeth and gear B has 32. They start with one particular pair of teeth touching. We need how many whole turns each makes before that same pair touches again.

Givens
  • Gear A has 48 teeth.
  • Gear B has 32 teeth.
  • They start meshed at a marked pair of teeth.
Unknowns
  • The number of full turns each gear makes before the marked pair meets again.
Constraints
  • Meshed gears pass teeth one for one, so both gears count the same number of teeth through the mesh point.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Count teeth, not turns. Each gear is back where it started after a multiple of its own tooth count, so the marked pair meets again at the first count that is a multiple of both.

3 · Execute3 carry out the plan

1See when each gear comes home

#9 Solve an Easier Related Problem 4.OA.B.4
Gear A is back at its starting position after every 48 teeth pass, gear B after every 32.
A:48,96, ⋯B:32,64,96, ⋯A: 48, 96,\ \cdots\quad B: 32, 64, 96,\ \cdots
Two lists of multiples, one per gear.

2Find the first shared count (the LCM)

#5 Look for a Pattern 6.NS.B.4
Both gears are home together at the first number on both lists.
lcm⁡(48,32)=96\operatorname{lcm}(48, 32) = 96
After 96 teeth the marked pair meets again.

3Turn the 96 teeth into full turns for each gear

#2 Make a Systematic List 4.OA.C.5
Divide the shared tooth count by each gear's own tooth count.
96÷48=2,96÷32=396 \div 48 = 2,\quad 96 \div 32 = 3
Gear A turns 2 times, gear B 3 times.
Answer: Gear A: 2 turns, Gear B: 3 turns
4 · Reviewdoes it hold up?

Both come to the same tooth count: 48 x 2 = 96 and 32 x 3 = 96. Gear B is the smaller one and turns the most, 3 times, which is what a smaller gear does.

Another way: Divide both tooth counts by their greatest common factor, 16: that gives 3 and 2, and the turns are those two numbers swapped.

Standardsmin grade 6
  • 6.NS.B.4 Find greatest common factor and least common multiple of two numbers — Finding 96 as the first tooth count both gears share.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing each gear's multiples of teeth.
  • 4.OA.C.5 Generate a number or shape pattern following a given rule — Converting the shared tooth count into turns for each gear.
💡Takeaway. Meshed gears count the same teeth, not the same turns -- so the smaller gear has to spin more to get back home.
Variant 12 hard answer: Gear A: 2 turns, Gear B: 3 turns

Two gears are meshed together: gear A has 5454 teeth and gear B has 3636 teeth. The gears begin turning together from the point where they are first meshed. To return to the position where the same two teeth mesh again, at least how many full turns must gear A and gear B each make?

Show solution
1 · Understandwhat's really being asked

Gear A has 54 teeth and gear B has 36. They start with one particular pair of teeth touching. We need how many whole turns each makes before that same pair touches again.

Givens
  • Gear A has 54 teeth.
  • Gear B has 36 teeth.
  • They start meshed at a marked pair of teeth.
Unknowns
  • The number of full turns each gear makes before the marked pair meets again.
Constraints
  • Meshed gears pass teeth one for one, so both gears count the same number of teeth through the mesh point.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#2 Make a Systematic List

Count teeth, not turns. Each gear is back where it started after a multiple of its own tooth count, so the marked pair meets again at the first count that is a multiple of both.

3 · Execute3 carry out the plan

1See when each gear comes home

#9 Solve an Easier Related Problem 4.OA.B.4
Gear A is back at its starting position after every 54 teeth pass, gear B after every 36.
A:54,108, ⋯B:36,72,108, ⋯A: 54, 108,\ \cdots\quad B: 36, 72, 108,\ \cdots
Two lists of multiples, one per gear.

2Find the first shared count (the LCM)

#5 Look for a Pattern 6.NS.B.4
Both gears are home together at the first number on both lists.
lcm⁡(54,36)=108\operatorname{lcm}(54, 36) = 108
After 108 teeth the marked pair meets again.

3Turn the 108 teeth into full turns for each gear

#2 Make a Systematic List 4.OA.C.5
Divide the shared tooth count by each gear's own tooth count.
108÷54=2,108÷36=3108 \div 54 = 2,\quad 108 \div 36 = 3
Gear A turns 2 times, gear B 3 times.
Answer: Gear A: 2 turns, Gear B: 3 turns
4 · Reviewdoes it hold up?

Both come to the same tooth count: 54 x 2 = 108 and 36 x 3 = 108. Gear B is the smaller one and turns the most, 3 times, which is what a smaller gear does.

Another way: Divide both tooth counts by their greatest common factor, 18: that gives 3 and 2, and the turns are those two numbers swapped.

Standardsmin grade 6
  • 6.NS.B.4 Find greatest common factor and least common multiple of two numbers — Finding 108 as the first tooth count both gears share.
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Listing each gear's multiples of teeth.
  • 4.OA.C.5 Generate a number or shape pattern following a given rule — Converting the shared tooth count into turns for each gear.
💡Takeaway. Meshed gears count the same teeth, not the same turns -- so the smaller gear has to spin more to get back home.