Find the smallest number which, when divided by 12,18 and 24 , leaves a remainder of 5 in each case.

Correct option is C

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 = $$2^2 × 3$$​​
18 = $$2 × 3^2$$​​
24 = $$2^3 × 3$$​​
LCM = $$2^3 × 3^2$$​ = 8 × 9 = 72
The required smallest number is LCM + Remainder.
Required number = 72 + 5 = 77
Final Answer
So the correct answer is (c)