Find the least number that leaves a remainder of 8 when divided by 12, 15, 20, and 54.

Given data

• N leaves remainder 8 upon division by 12, 15, 20, and 54.

Concept / Approach

• N − 8 must be divisible by all four numbers ⇒ N − 8 is a multiple of their L.C.M.

Step-by-step calculation

Prime factors: 12 = 2^2×3, 15 = 3×5, 20 = 2^2×5, 54 = 2×3^3LCM = 2^2 × 3^3 × 5 = 4 × 27 × 5 = 540Smallest N = 540 + 8 = 548

Verification

548 − 8 = 540 is divisible by 12, 15, 20 and 54.

Common pitfalls

• Answering 540 (forgetting to add back the remainder 8).

Final Answer

Required number = 548.