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

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

A.

101

B.

65

C.

77

D.

125

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 = 22×32^2 × 3​​
18 = 2×322 × 3^2​​
24 = 23×3 2^3 × 3​​
LCM = 23×322^3 × 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)

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