Discussion

Answer: C

The problem asks for a number N such that \(N \equiv 8 \pmod{12}\), \(N \equiv 8 \pmod{15}\), etc. This means \(N-8\) must be a multiple of 12, 15, 20, and 54. So, \(N-8\) must be a multiple of their LCM.

First, find the LCM of 12, 15, 20, 54.

\(12 = 2^2 \times 3\)

\(15 = 3 \times 5\)

\(20 = 2^2 \times 5\)

\(54 = 2 \times 3^3\)

LCM = \(2^2 \times 3^3 \times 5 = 4 \times 27 \times 5 = 540\).

So, \(N-8\) must be a multiple of 540. For the least such number, \(N-8 = 540\). Therefore, \(N = 540 + 8 = 548\).