For positive integer k, is the expression (k + 2)(k2 + 4k + 3) divisible by 4?
(1) k is divisible by 8.
(2) | k + 1 3 | is an odd integer. |
Solution
The quadratic expression k2 + 4k + 3 can be factored to yield (k + 1)(k + 3). Thus, the expression in the question stem can be restated as (k + 1)(k + 2)(k + 3), or the product of three consecutive integers. This product will be divisible by 4 if one of two conditions are met:
If k is odd, both k + 1 and k + 3 must be even, and the product (k + 1)(k + 2)(k + 3) would be divisible by 2 twice. Therefore, if k is odd, our product must be divisible by 4.
If k is even, both k + 1 and k + 3 must be odd, and the product (k + 1)(k + 2)(k + 3) would be divisible by 4 only if k + 2, the only even integer among the three, were itself divisible by 4.
The question might therefore be rephrased “Is k odd, OR is k + 2 divisible by 4?” Note that a ‘yes’ to either of the conditions would suffice, but to answer 'no' to the question would require a ‘no’ to both conditions.
(1) SUFFICIENT: If k is divisible by 8, it must be both even and divisible by 4. If k is divisible by 4, k + 2 cannot be divisible by 4. Therefore, statement (1) yields a definitive ‘no’ to both conditions in our rephrased question; k is not odd, and k + 2 is not divisible by 4.
(2) INSUFFICIENT: If k + 1 is divisible by 3, k + 1 must be an odd integer, and k an even integer. However, we do not have sufficient information to determine whether k or k + 2 is divisible by 4.
The correct answer is A.
Query
I think the answer is D. In Statement (2), for (k+1)/3 to be an odd INTEGER, k+1 has to be divisible by 3. Also, k must be even. Therefore, k could be 2,8,... And if k is an even integer, (k + 1)(k + 2)(k + 3) will not be divisible by 4 for any of these values of k. Therefore, Statement (2) also gives us the answer. Please tell me where I'm going wrong.
1 comment:
Put 2 as a solution for k in (k+1)(k+2)(k+3). This equation is this divisible by 4. For other solutions, k=8,14 etc, this is not the case. Hence, (2) is Insufficient.
Post a Comment