Discussion
The least number which when divided by the numbers 3, 5, 6, 8, 10 and 12 leaves in each case a remainder 2 but which when divided by 13 leaves no remainder?
Answer: B
A number leaving the same remainder 2 under all six divisors is 2 more than a common multiple of them, so start from their lowest common multiple:
\[ \operatorname{LCM}(3,5,6,8,10,12) = 120 \quad\Longrightarrow\quad N = 120k + 2 \]
Now impose divisibility by 13. Since \( 120 = 9(13) + 3 \), we have \( 120 \equiv 3 \pmod{13} \), so
\[ 3k + 2 \equiv 0 \pmod{13} \quad\Longrightarrow\quad 3k \equiv 11 \pmod{13} \]
Multiplying by 9, the inverse of 3 modulo 13 because \( 3 \times 9 = 27 \equiv 1 \):
\[ k \equiv 99 \equiv 8 \pmod{13} \quad\Longrightarrow\quad N = 120(8) + 2 = 962 \]
Check: 962 divided by 3, 5, 6, 8, 10 and 12 leaves 2 every time, and \( 962 = 13 \times 74 \) exactly. Of the distractors, 1562 clears the remainder test but is not a multiple of 13, and 1586 is a multiple of 13 but leaves remainder 1 on division by 5.
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