reto wrote:
tarek99 wrote:
If n and k are positive integers, is n divisible by 6?
(1) n = k(k + 1)(k - 1)
(2) k – 1 is a multiple of 3
Here's my way to solve this:Statement (1):Rewrite n = k(k + 1)(k - 1) to: \(k^3-k\) and plug in various positive integers to see that the result will alsways be divisible by 3. Therefore AC 1 = Sufficient
Statement (2)Gives you just an idea about the term k-1, but nothing about k or k+1 itself. Therefore clearly insufficient.
reto, a couple of points about your solution (although you were able to get to the correct answer):
For statement 1, k(k-1)(k+1) will ALWAYS be divisible by 6. Think of this way. For any number to be divisible by 6, it needs to be divisible by
BOTH 2 and 3 at the same time.
Consider 2 cases, if k =2p, then k+1 or k-1 MUST be disivible by 3. Thus the product k(k-1)(k+1) will always be divisible by 6.
Case 2: if k = 3p, then k-1 or k+1 MUST be divisble by 2. Thus the product k(k-1)(k+1) will always be divisible by 6.
As for your interpretation of statement 2, we do not need to even look at the statement and start computing numbers as the question stem asks us whether "n" is divisible by 6 and NOT whether "k" is divisible by 6. Statement 2 does not provide any relation between k and n and hence straightaway you can reject this statement.
Also you are incorrect to say that as k-1=3p, you have no information about k+1 or k. This is not correct. If k-1=3p., then 1 of k or k+1 MUST be divisible by 2 (you can check with some numbers). Thus you do get some actionable information (albeit not pertinent to the original question asked!).
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