Find the smallest number which, when divided by 12,18 and 24 , leaves a remainder of 5 in each case.
Given:
Divisors = 12, 18, 24
Remainder in each case = 5
Formula Used:
Required number = LCM(a, b, c) + Remainder
Solution:
First, find the LCM of 12, 18, and 24.
Prime factorization:
12 =
18 =
24 =
LCM = = 8 × 9 = 72
The required smallest number is LCM + Remainder.
Required number = 72 + 5 = 77
Final Answer
So the correct answer is (c)
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