HCF Problem 1

Three company of soldiers containing 120, 192, and 144 soldiers are to be broken down into smaller groups such that each group contains soldiers from one company only and all the groups have equal number of soldiers. What is the least number of total groups formed?

Which one to go LCM or HCF??

From the problem, it is clearly mentioned least number of total groups. Total groups can be least only when the soldiers in each group are more. Also we should make sure each group to have equal number of soldiers from one company only. To satisfy both the statements above we have to find the largest group that is common to all the 3 companies. So we will go for HCF of 120, 192 and 144.


HCF of 120, 192 and 144 is 24.

How to do HCF of 120,192,144?

Lets do HCF with Division Method:

Get the largest and smallest number from the given list which is 192 and 120.

192 Divided by 120 leaves Remainder 72
120 Divided by 72  leaves Remainder  48
72  Divided by 48 leaves Remainder 24
48  Divided by 24 leaves Remainder 0.

Therefor 24 is the Highest common factor which divides 192,144 and 120 equally.     

Now divide each group by 24 to get number of groups  

120 / 24 = 5 
192 / 24 = 8
144 / 24 = 6

Total number of least groups having equal soldiers from 3 companies is 5 + 8 + 6 = 19. 

Problems on LCM

Now Lets take a look at  below problem.

1. Ram takes 3 mins to complete one round on a circular track. Raghu takes 6 mins to completed one round on the same circular track. At what time Ram and Raghu cross each other on the track?

Sol:

Ram    takes 3 mins to complete one round.
Raghu takes 6 mins to complete one round.

Consider dotted lines as time.
Ram    ............................... (1st round Completed in 3 mins)
Raghu ............................... (1/2 round Completed in 3 mins)

Ram   ...............................(2nd round Completed in 6 mins)
Raghu ...............................(1st round Completed in 6 mins)

That means , Ram and Raghu cross each other in 6 mins.

6 mins is nothing but the least time at which Ram and Raghu cross eachother for first time on the circular track.
LCM (3,6) is 6.

2.) How many times both Ram and Raghu cross each other in one hour?

Sol: Since for every 6mins they cross each other one time.
For 60 mins (1 hour) they cross 10 times.

LCM and GCD Continued...

LCM of  8 and 12 can be found as shown below.

LCM (8,12) = 8 x 12             8 x 12
                      ------------ = ---------=  24.
                      GCD(8,12)          4

GCD of 8 and 12 can be found as show below.

12 divided by 8 leaves remainder 4
  8 divided by 4 leaves remainder 0.
There GCD of 8 and 12 is 4.

LCM and GCD

LCM : Least Common Multiple
GCD: Greatest Common Divisor

What is Multiple: A number multiplied by another number.
8 x 1 =   8
8 x 2 = 16
8 x 3 = 24
8 x 4 = 32
8 x 5 = 40
.
.
8,16,24,32,40 are multiples of 8.

What is Divisor: A number which can divide another number leaving no remainder.
 8 Divided by 2 leaves 0 as remainder.
 Therefore, 2 is Divisor.

Now lets see in detail LCM:
What is the LCM of 8 and 12.

From the example below even though 24,48,... etc are common multiples between 8 and 12. The least one among them is 24.

Therefore, LCM of 8 and 12 is 24.

8 x 1 = 8
8 x 2 = 16
8 x 3 = 24
8 x 4 = 32
8 x 5 = 40
8 x 6 = 48

12 x 1 =12
12 x 2 = 24
12 x 3 = 36
12 x 4 = 48
12 x 5 = 60

Now lets see in detail GCD: 
What is the GCD of 8 and 12.

Divisors of 8 are 1,2,4,8. All these will leave remainder = 0.
Divisors of 12 are 1,2,3,4,6,12. All these will leave remainder = 0

Therefore, GCD of 8 and 12 is 4.