Discussion
Find the least number which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8.
Answer: C
Correct answer: 548
A number leaving the same remainder 8 under every one of these divisors has the form (a common multiple of them) \(+\ 8\). The least such number uses the LCM.
\[12 = 2^2\times 3,\quad 15 = 3\times 5,\quad 20 = 2^2\times 5,\quad 54 = 2\times 3^3\]
Taking the highest power of each prime, \(\text{LCM} = 2^2\times 3^3\times 5 = 540\).
\[N = 540 + 8 = 548\]
Strictly, 8 itself also leaves remainder 8 under each of these divisors, since it is smaller than all of them. The intended reading, standard for this question type, is the least number larger than the divisors, which is 548.
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