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If n is a positive integer, is n^3 – n divisible by 4 ?
(1) n = 2k + 1, where k is an integer
(2) n^2 + n is divisible by 6
Solution:We need to determine whether n^3 - n is divisible by 4. Notice that n^3 - n can be factored as n(n^2 - 1) = n(n - 1)(n + 1), which can be expressed as a product of three consecutive integers: (n - 1)(n)(n + 1).
Statement One Alone:
n = 2k + 1, where k is an integer
Statement one tells us that n is an odd integer. Thus, (n - 1) and (n + 1) are both even. We have (n - 1)(n)(n + 1) as even x odd x even, and so each even factor contributes at least one 2 to the product. Therefore, n^3 – n is divisible by 4.
(Note that even if n = 1, resulting in a product of 0, statement one still holds because zero is divisible by 4.)
Statement one alone is sufficient.
Statement Two Alone:n^2 + n is divisible by 6
If n = 2, we see that n^2 + n = 4 + 2 = 6 is divisible by 6. However, n(n - 1)(n + 1) = 2(1)(3) = 6 is not divisible by 4.
On the other hand, if n = 3, then n^2 + n = 9 + 3 = 12 is also divisible by 6. In this case, n(n - 1)(n + 1) = 3(2)(4) IS divisible by 4.
Statement two alone is not sufficient.
Answer: A