CLASS-8
PROBLEM & 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.
Answer.)
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.)