# CLASS-8PROBLEM & SOLUTION OF HCF & LCM

PROBLEM & SOLUTION

Example.1)  Find the greatest number that will divide 325, 725 and 1685 leaving the reminders such as 13, 17, and 9 respectively.

Answer)  as per the given condition,  325 – 13 = 312,

725 – 17 = 708,

1685 – 9 =  1676

The required number is the HCF of 312, 708, 1676. First of all, we will consider 708 as Divisor and 1676 as a Dividend and will find the HCF of 708 & 1676

708 ) 1676 ( 2

1416

-----------

260 ) 708 ( 2

520

-----------

188 ) 260 ( 1

188

-----------

72 ) 188 ( 2

144

----------

44 ) 72 ( 1

44

---------

28 ) 44 ( 1

28

--------

16 ) 28 ( 1

16

-------

12 ) 16 ( 1

12

-------

4 ) 12 ( 3

12

---------

0

So, the HCF of 708 & 1676 is 4, now we have to find out the HCF of 4 & 312. As per the rules, we will consider 4 as Divisor and 312 as Dividend.

4 ) 312 ( 78

28

--------

32

32

---------

0

So, HCF of  312, 708, 1676 is 4         (Ans.)

Example.2) find the greatest number of five digits which is exactly divisible by 12, 25, 18, 50 and 75

Answer) First, we have to find the LCM of given number 12, 25, 18, 50 and 75

LCM = 2 X 3 X 25 X 2 X 3 = 900

The greatest number of 5 digits should be 99999

Let, we divide this by the LCM

900 )  99999 ( 111

900

----------

999

900

-----------

999

900

------------

99

So, 99999 ÷ LCM leaves the remainder 99

Hence the required number = 99999 – remainder

=  99999 – 99

= 99900                  (Ans.)

Example.3) Five bells start ringing together, if the bells ring at intervals of 18, 56, 72, 96, & 120 seconds respectively, after what interval of time will they ring together again ?

Answer) The required time (in seconds) = the LCM of  18, 56, 72, 96, 120

The required LCM -

=  2 X 3 X 2 X 2 X 3 X 7 X 4 X 5

=  10080

Hence the required time is  = 10080 seconds =  168 minutes = 2 hours 48 minutes

So, after 2 hours 48 minutes, all the five bells will ring together.     (Ans.)

Example.4)

a) Find the LCM of 36, 42, 72, 108

b) find the smallest number which when divided by  36, 42, 72, & 108 leave the remainder 3 in each case.

c) find the smallest number which when increased by 7 is exactly divisible by 36, 42, 72, & 108.

a)  as per the given condition first we have to find out the LCM of 36, 42, 72, & 108.

So, LCM of  36, 42 , 72 & 108 is

=  2 X 2 X 3 X 3 X 7 X 2 X 3 X 1

= 1512

b)  according to the given condition the required number would be –

LCM + 3 = 1512 + 3 =  1515

c) according to the given condition the required number would be –

LCM – 7 = 1512 – 7

= 1505               (Ans.)

Example.5)  find the smallest number of Six digits which is exactly divisible by  12, 24, 56 & 72

Answer.) According to the given condition first, we have to find out the LCM of  12, 24, 56 & 72.

LCM of 12, 24, 56 & 72 is

=  2 X 2 X 3 X 2 X 7 X 3

=  504

As per the condition six-digit smallest number is = 100000

Let us divide this by the obtained LCM, so

504 ) 100000 ( 198

504

-----------

4960

4536

-----------

4240

4032

-----------

208

So, the 198th multiple of 504 is less than 100000

So, the required number is = 199th multiple of 504

= 199 X 504 =  100296

Or, the required number is =  100000 – Reminder + LCM

=  100000 – 208 + 504

=  100000 + 296

=  100296              (Ans.)

Example.6) The LCM of two numbers is 15 times of their HCF and the difference between the LCM and the HCF is 280. If one of the numbers is 50, then find the others.

Answer) Let the HCF of the two number is  = z

The LCM of the two numbers = 15z

As per the given condition LCM – the HCF = 280

So,     15z – z  =  280

14 z  = 280,

Z  = 280 / 14

=  20

So, according to the given condition if the HCF is = 20

Then, LCM is  = 20 X 15 = 300

As we know that, the product of two numbers =  HCF X LCM

If one product is 50, then let another product is Y

Now,  50 X Y =  20 X 300

Y = 6000 / 50  =  120

So, the desired number is  120.             (Ans.)