Official Solution:If \(a\), \(b\), and \(c\) are positive integers, is \(a*b*c\) divisible by 24? (1) \(a\), \(b\), and \(c\) are consecutive even integers.
Given: \(a=2k-2\), \(b=2k\), and \(c=2k+2\) for some integer \(k\). So, \(abc = (2k-2)2k(2k+2) = 8(k-1)k(k+1)\). Now, \((k-1)\), \(k\), and \((k+1)\) are 3 consecutive integers, which means that one of them must be a multiple of 3. Therefore, \(abc\) is divisible by both 8 and 3, and consequently by 24. Sufficient.
Or even without the formulas: the product of 3 consecutive even integers will have \(2*2*2=8\) as a factor. Plus, out of 3 consecutive even integers, one must be a multiple of 3. Thus, \(abc\) is divisible by both 8 and 3, so by 24 too.
(2) \(a*b\) is divisible by 12.
This is clearly insufficient as we don't have any information about \(c\) If \(ab=12\) and \(c=1\), the answer is NO. However, if \(ab=24\) and \(c\) is any integer, the answer is YES. Not sufficient.
Answer: A