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From 1) we know \(\frac{n}{12} = x + \frac{1}{12}\) so adjusting it to become \(n = 12x + 1\) if we let \(x = 0\) then\(n = 1\), plug in in the question stem \(\frac{1}{6}\) remainder is 1, if we let \(x = 1\), then \(n = 13\), plug in \(\frac{13}{6} = 2 \frac{1}{6}\) another remainder of 1 and so it goes. A is sufficient.

From 2) \(\frac{n}{3} = x + \frac{1}{3}\) so adjusting it to become \(n = 3x + 1\) if we let \(x = 0\) then \(n = 1\), plug in in the question stem \(\frac{1}{6}\) remainder is 1, if we let \(x = 1\), then \(n = 4\), plug in question stem \(\frac{4}{6}\) a remainder of 4. Because we get two different answers it is insufficient.

The answer is A
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From statement I alone, the remainder is 1 when the positive integer is divided by 12.

Therefore, n = 12k + 1, using the basic division algorithm
All multiples of 12 are necessarily divisible by 6 also. Therefore, when n is divided by 6, the remainder will be 1 since 12k is divisible by 6.

Statement I alone is sufficient to answer the question. Answer options B, C and E can be eliminated. Possible answer options are A or D.

From statement II alone, the remainder is 1 when the positive integer is divided by 3.

Therefore, n = 3k + 1.
If k = even, 3k will be a multiple of 6 and so the remainder will be 1.
If k = odd, 3k will be an odd multiple of 3 and the remainder will therefore increase by 3 i.e. the remainder will be 4.

Statement II alone is not sufficient to answer the question. Answer option D can be eliminated.

The correct answer option is A.

Hope that helps!
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Every multiple of 12 is a multiple of 6. So if, from Statement 1, n is 1 greater than a multiple of 12, then n must be 1 greater than a multiple of 6, and the remainder is 1 when we divide n by 6. So Statement 1 is sufficient.

Statement 2 is not sufficient alone, since n can equal 1 or 4, and those give remainders of 1 and 4 respectively when we divide them by 6.
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