Practice

Question 1

1. Cabin 12 goes at every multiple of 6 minutes, and Cabin 25 at every multiple of 16.Cabin 12: 6, 12, 18, 24, 30, 36, 42, 48
2. They can only meet at a minute that is on both lists — a common multiple.Cabin 25: 16, 32, 48
3. The first time they meet is the smallest number on both lists, so break each interval into its prime factors.6 = 2 × 3, 16 = 2 × 2 × 2 × 2
4. Take every factor as many times as it appears in either one.LCM = 2 × 2 × 2 × 2 × 3 = 48
5. So they meet again after 48 minutes — 8 goes of Cabin 12 and 3 of Cabin 25.8 × 6 = 48 = 3 × 16
6. Count that on from 8.15 a.m..8.15 a.m. + 48 min = 9.03 a.m.

Cabin 12 and cabin 25 are next at the platform together at 9.03 a.m.

The pattern. They can only meet at a minute that is on both lists of multiples. The first such minute is the lowest common multiple: 2 × 2 × 2 × 2 × 3 = 48.

Question 2

1. Washing machine 1 goes at every multiple of 20 minutes, and Washing machine 2 at every multiple of 35.Washing machine 1: 20, 40, 60, 80, 100, 120, 140
2. They can only meet at a minute that is on both lists — a common multiple.Washing machine 2: 35, 70, 105, 140
3. The first time they meet is the smallest number on both lists, so break each interval into its prime factors.20 = 2 × 2 × 5, 35 = 5 × 7
4. Take every factor as many times as it appears in either one.LCM = 2 × 2 × 5 × 7 = 140
5. So they meet again after 140 minutes — 7 goes of Washing machine 1 and 4 of Washing machine 2.7 × 20 = 140 = 4 × 35
6. That is the answer the question asks for.140 minutes

The two washing machines next finish a wash together after 140 minutes.

The pattern. They can only meet at a minute that is on both lists of multiples. The first such minute is the lowest common multiple: 2 × 2 × 5 × 7 = 140.

Question 3

1. Robot vacuum 1 goes at every multiple of 18 minutes, and Robot vacuum 2 at every multiple of 21.Robot vacuum 1: 18, 36, 54, 72, 90, 108, 126
2. They can only meet at a minute that is on both lists — a common multiple.Robot vacuum 2: 21, 42, 63, 84, 105, 126
3. The first time they meet is the smallest number on both lists, so break each interval into its prime factors.18 = 2 × 3 × 3, 21 = 3 × 7
4. Take every factor as many times as it appears in either one.LCM = 2 × 3 × 3 × 7 = 126
5. So they meet again after 126 minutes — 7 goes of Robot vacuum 1 and 6 of Robot vacuum 2.7 × 18 = 126 = 6 × 21
6. That is the answer the question asks for.126 minutes

The two robot vacuums are next at the dock together after 126 minutes.

The pattern. They can only meet at a minute that is on both lists of multiples. The first such minute is the lowest common multiple: 2 × 3 × 3 × 7 = 126.

Question 4

1. Robot arm 1 goes at every multiple of 9 minutes, and Robot arm 2 at every multiple of 15.Robot arm 1: 9, 18, 27, 36, 45
2. They can only meet at a minute that is on both lists — a common multiple.Robot arm 2: 15, 30, 45
3. The first time they meet is the smallest number on both lists, so break each interval into its prime factors.9 = 3 × 3, 15 = 3 × 5
4. Take every factor as many times as it appears in either one.LCM = 3 × 3 × 5 = 45
5. So they meet again after 45 minutes — 5 goes of Robot arm 1 and 3 of Robot arm 2.5 × 9 = 45 = 3 × 15
6. That is the answer the question asks for.45 minutes

The two robot arms next complete a cycle together after 45 minutes.

The pattern. They can only meet at a minute that is on both lists of multiples. The first such minute is the lowest common multiple: 3 × 3 × 5 = 45.

Question 5

1. Sprinkler A goes at every multiple of 21 minutes, and Sprinkler B at every multiple of 24.Sprinkler A: 21, 42, 63, 84, 105, 126, 147, 168
2. They can only meet at a minute that is on both lists — a common multiple.Sprinkler B: 24, 48, 72, 96, 120, 144, 168
3. The first time they meet is the smallest number on both lists, so break each interval into its prime factors.21 = 3 × 7, 24 = 2 × 2 × 2 × 3
4. Take every factor as many times as it appears in either one.LCM = 2 × 2 × 2 × 3 × 7 = 168
5. So they meet again after 168 minutes — 8 goes of Sprinkler A and 7 of Sprinkler B.8 × 21 = 168 = 7 × 24
6. Count that on from 7.30 a.m..7.30 a.m. + 168 min = 10.18 a.m.

The two sprinklers next switch on together at 10.18 a.m.

The pattern. They can only meet at a minute that is on both lists of multiples. The first such minute is the lowest common multiple: 2 × 2 × 2 × 3 × 7 = 168.

Question 6

1. The east fountain goes at every multiple of 15 minutes, and The west fountain at every multiple of 35.The east fountain: 15, 30, 45, 60, 75, 90, 105
2. They can only meet at a minute that is on both lists — a common multiple.The west fountain: 35, 70, 105
3. The first time they meet is the smallest number on both lists, so break each interval into its prime factors.15 = 3 × 5, 35 = 5 × 7
4. Take every factor as many times as it appears in either one.LCM = 3 × 5 × 7 = 105
5. So they meet again after 105 minutes — 7 goes of The east fountain and 3 of The west fountain.7 × 15 = 105 = 3 × 35
6. Count that on from 4.00 p.m..4.00 p.m. + 105 min = 5.45 p.m.

The two fountains next spurt together at 5.45 p.m.

The pattern. They can only meet at a minute that is on both lists of multiples. The first such minute is the lowest common multiple: 3 × 5 × 7 = 105.

Question 7

1. The red beacon goes at every multiple of 9 minutes, and The green beacon at every multiple of 21.The red beacon: 9, 18, 27, 36, 45, 54, 63
2. They can only meet at a minute that is on both lists — a common multiple.The green beacon: 21, 42, 63
3. The first time they meet is the smallest number on both lists, so break each interval into its prime factors.9 = 3 × 3, 21 = 3 × 7
4. Take every factor as many times as it appears in either one.LCM = 3 × 3 × 7 = 63
5. So they meet again after 63 minutes — 7 goes of The red beacon and 3 of The green beacon.7 × 9 = 63 = 3 × 21
6. Count that on from 7.30 a.m..7.30 a.m. + 63 min = 8.33 a.m.

The two beacons next flash together at 8.33 a.m.

The pattern. They can only meet at a minute that is on both lists of multiples. The first such minute is the lowest common multiple: 3 × 3 × 7 = 63.

Question 8

1. The red beacon goes at every multiple of 15 minutes, and The green beacon at every multiple of 35.The red beacon: 15, 30, 45, 60, 75, 90, 105
2. They can only meet at a minute that is on both lists — a common multiple.The green beacon: 35, 70, 105
3. The first time they meet is the smallest number on both lists, so break each interval into its prime factors.15 = 3 × 5, 35 = 5 × 7
4. Take every factor as many times as it appears in either one.LCM = 3 × 5 × 7 = 105
5. So they meet again after 105 minutes — 7 goes of The red beacon and 3 of The green beacon.7 × 15 = 105 = 3 × 35
6. Count that on from 4.00 p.m..4.00 p.m. + 105 min = 5.45 p.m.

The two beacons next flash together at 5.45 p.m.

The pattern. They can only meet at a minute that is on both lists of multiples. The first such minute is the lowest common multiple: 3 × 5 × 7 = 105.

Question 9

1. Lift A goes at every multiple of 6 minutes, and Lift B at every multiple of 9.Lift A: 6, 12, 18
2. They can only meet at a minute that is on both lists — a common multiple.Lift B: 9, 18
3. The first time they meet is the smallest number on both lists, so break each interval into its prime factors.6 = 2 × 3, 9 = 3 × 3
4. Take every factor as many times as it appears in either one.LCM = 2 × 3 × 3 = 18
5. So they meet again after 18 minutes — 3 goes of Lift A and 2 of Lift B.3 × 6 = 18 = 2 × 9
6. Count that on from 7.00 a.m..7.00 a.m. + 18 min = 7.18 a.m.

The two lifts are next on the ground floor together at 7.18 a.m.

The pattern. They can only meet at a minute that is on both lists of multiples. The first such minute is the lowest common multiple: 2 × 3 × 3 = 18.

Question 10

1. Lift A goes at every multiple of 3 minutes, and Lift B at every multiple of 7.Lift A: 3, 6, 9, 12, 15, 18, 21
2. They can only meet at a minute that is on both lists — a common multiple.Lift B: 7, 14, 21
3. The first time they meet is the smallest number on both lists, so break each interval into its prime factors.3 = 3, 7 = 7
4. Take every factor as many times as it appears in either one.LCM = 3 × 7 = 21
5. So they meet again after 21 minutes — 7 goes of Lift A and 3 of Lift B.7 × 3 = 21 = 3 × 7
6. That is the answer the question asks for.21 minutes

The two lifts are next on the ground floor together after 21 minutes.

The pattern. They can only meet at a minute that is on both lists of multiples. The first such minute is the lowest common multiple: 3 × 7 = 21.

Question 11

1. Cable car 4 goes at every multiple of 14 minutes, and Cable car 9 at every multiple of 16.Cable car 4: 14, 28, 42, 56, 70, 84, 98, 112
2. They can only meet at a minute that is on both lists — a common multiple.Cable car 9: 16, 32, 48, 64, 80, 96, 112
3. The first time they meet is the smallest number on both lists, so break each interval into its prime factors.14 = 2 × 7, 16 = 2 × 2 × 2 × 2
4. Take every factor as many times as it appears in either one.LCM = 2 × 2 × 2 × 2 × 7 = 112
5. So they meet again after 112 minutes — 8 goes of Cable car 4 and 7 of Cable car 9.8 × 14 = 112 = 7 × 16
6. Count that on from 9.00 a.m..9.00 a.m. + 112 min = 10.52 a.m.

The two cable cars next leave together at 10.52 a.m.

The pattern. They can only meet at a minute that is on both lists of multiples. The first such minute is the lowest common multiple: 2 × 2 × 2 × 2 × 7 = 112.

Question 12

1. Bus 58 goes at every multiple of 21 minutes, and Bus 71 at every multiple of 28.Bus 58: 21, 42, 63, 84
2. They can only meet at a minute that is on both lists — a common multiple.Bus 71: 28, 56, 84
3. The first time they meet is the smallest number on both lists, so break each interval into its prime factors.21 = 3 × 7, 28 = 2 × 2 × 7
4. Take every factor as many times as it appears in either one.LCM = 2 × 2 × 3 × 7 = 84
5. So they meet again after 84 minutes — 4 goes of Bus 58 and 3 of Bus 71.4 × 21 = 84 = 3 × 28
6. Count that on from 7.30 a.m..7.30 a.m. + 84 min = 8.54 a.m.

The two buses next leave together at 8.54 a.m.

The pattern. They can only meet at a minute that is on both lists of multiples. The first such minute is the lowest common multiple: 2 × 2 × 3 × 7 = 84.

Question 13

1. Bus 58 goes at every multiple of 20 minutes, and Bus 71 at every multiple of 30.Bus 58: 20, 40, 60
2. They can only meet at a minute that is on both lists — a common multiple.Bus 71: 30, 60
3. The first time they meet is the smallest number on both lists, so break each interval into its prime factors.20 = 2 × 2 × 5, 30 = 2 × 3 × 5
4. Take every factor as many times as it appears in either one.LCM = 2 × 2 × 3 × 5 = 60
5. So they meet again after 60 minutes — 3 goes of Bus 58 and 2 of Bus 71.3 × 20 = 60 = 2 × 30
6. Count that on from 7.30 a.m..7.30 a.m. + 60 min = 8.30 a.m.

The two buses next leave together at 8.30 a.m.

The pattern. They can only meet at a minute that is on both lists of multiples. The first such minute is the lowest common multiple: 2 × 2 × 3 × 5 = 60.

Question 14

1. Ride A goes at every multiple of 15 minutes, and Ride B at every multiple of 24.Ride A: 15, 30, 45, 60, 75, 90, 105, 120
2. They can only meet at a minute that is on both lists — a common multiple.Ride B: 24, 48, 72, 96, 120
3. The first time they meet is the smallest number on both lists, so break each interval into its prime factors.15 = 3 × 5, 24 = 2 × 2 × 2 × 3
4. Take every factor as many times as it appears in either one.LCM = 2 × 2 × 2 × 3 × 5 = 120
5. So they meet again after 120 minutes — 8 goes of Ride A and 5 of Ride B.8 × 15 = 120 = 5 × 24
6. Count that on from 7.00 a.m..7.00 a.m. + 120 min = 9.00 a.m.

The two rides next start a round together at 9.00 a.m.

The pattern. They can only meet at a minute that is on both lists of multiples. The first such minute is the lowest common multiple: 2 × 2 × 2 × 3 × 5 = 120.