Question

What is the smallest number which when divided by 24, 36 and 54 gives a remainder of 5 each time?

Answer 1

We have to find prime factorisation of 24, 36, and 54.

Prime factorisation of 24 = 2 × 2 × 2 × 3
Prime factorisation of 36 = 2 × 2 × 3 × 3
Prime factorisation of 54 = 2 × 3 × 3 × 3
∴ Required LCM = 2 × 2 × 2 × 3 × 3 × 3 = 216

Thus, 216 is the smallest number exactly divisible by 24, 36, and 54.
To get the remainder as 5:
Smallest number = 216 + 5 = 221
Thus, the required number is 221.