Problem S. Find the least number which when divided by 20,30 and 4
of 7 in each ease.
Solution: Clearly, the least number that is divisible by 20,30 und 455
multiplier of given numbers.
Now, the LCM of 20, 30 and 40
-2x2x2x3 x 5
120
120 is a number which is exactly divisible by
given number
Now, we have to find a number that leaves a remainder
of 7 in each case. So, the required number is 7 more than
120 i.e., 127.
EXERCISE 3.5
1. Using prime factorisation method, find the L.C.M. of:
Answer:
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