Shortcut - HCF and LCM
Finding both HCF and LCM using single L-Division: While doing the L-Division method, initially divide only using the prime numbers which are common factors for all the numbers, then divide using the prime numbers which are not common.
Question:
Find LCM, and HCF for the numbers 240, 450 and 570.
Answer:2 240, 450, 570
5 120, 225, 285
3 24, 45, 57
19 8, 15, 19
5 8, 15, 1
3 8, 3, 1
2 8, 1, 1
2 4, 1 , 1
2 2, 1, 1
1, 1, 1
The common factors for the three numbers are 2, 3 and 5.
Variation between common factors and the others is shown by using different font.
Multiply only the common factors to get the HCF
HCF = 2 x 3 x 5 = 30
Multiply all the factors to get LCM
LCM= 2 x 3 x 5 x 19 x 5 x 3 x 2 x 2 x 2
= $2^{4} \times 3^{2}\times 5^{2} \times 19^{1}$
= 68400
Shortcut - HCF and LCM
Application problem using HCF. There will be different sets of elements with different quantities in each.
They have to be arranged based on the following conditions.
1. Every group should have equal number of elements.
2. Every group should have same kind of elements.
Under these conditions one must find out:
1. The maximum number of elements per group.
2. The minimum number of groups required.
HCF of the numbers of elements in all the groups will give the maximum number of elements per group.
Total number of elements divided by HCF will give the minimum number of groups required.
Question:
There are 40 apples and 32 oranges. They have to be packed in boxes such that each box will have one kind of fruit and every box will have equal number of fruits. Under this condition, what is the:
1. Maximum number of fruits per box?
2. Minimum number of boxes required?
Answer: Maximum number of fruits per box = HCF(40, 32) = 8
Minimum number of boxes required = (40 + 32)/8 = 9
Shortcut - HCF and LCM
Application problem using HCF:
For three or more numbers, the HCF of the difference between the successive numbers arranged in ascending order is the largest number that can divide the given numbers and leave same remainder.
Question:
What is the largest number that can divide 212, 254, 310 and 338 and leave the same remainder?
Answer: The difference between the numbers = 42, 56 and 28.
HCF (42, 56, 28) = 14
Remainder (212/14) = 2
Remainder (254/14) = 2
Remainder (310/14) = 2
Remainder (338/14) = 2
14 is the largest number that can divide the given numbers and leave the same remainder.
Question:
What is the largest number that can divide 34, 52 and 88 and leave the same remainder?
Answer: Difference between the numbers = 18 and 36
HCF (18, 36) = 18
18 is the largest number that can divide 34, 52 and 88 and leave same remainder.
Shortcut - HCF and LCM
Application problem using LCM:
When different events occur in different intervals of their own and if all the events are started at the same time, at some point of time all the events will again occur together. To find the duration between two common occurrences, LCM of the intervals of each event has to be calculated.
Question:
There are three different bells. The first bell rings every 4 hour, the second bell rings every 6 hours and the third bell rings every 15 hours. If all the three bells are rung at same time, how long will it take for them to ring together again at the same time?
Answer: Take LCM of 4, 6, 15
LCM (4,6,15) = 60
All the three bells will ring after 60 hours.
Question:
There are three different alarms. The first alarm rings every 2 minutes, the second alarm rings every 3 minutes and the third alarm rings every 5 minutes. If all the three alarms are rung at 6 am together, at what time the three alarms will ring together for the next time?
Answer:LCM (2, 3, 5) = 30. After 30 minutes = 06:30 am.