Data Sufficiency Question - I can't find my error

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Hey guys,

I have a question on this D.S. problem, can someone explain why my solution is incorrect?

If k is a positive integer and n = k(k + 7), is n divisible by 6?

1. K is odd

Test cases:
K=1 n=1(1+7) = 8, is 8/6 NO
K=3 n=3(3+7)=30, is 30/6 YES
INSUFFICIENT

2. When k is divided by 3, the remainder is 2

Test Cases:
K=1 1/3=0 remainder 2, n=1(1+7) = 8, is 8/6 NO
K=5 5/3 =1 remainder 2, n=5(5+7) =60, is 60/6 YES
INSUFFICIENT

Combined:
K=1 overlaps - NO
K=5 overlaps - YES

OA - B

I've seen a complex math approach that seems to prove that B is sufficient but why would the test cases produce a different result? What am I missing here?

Thanks in advance.

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by theCodeToGMAT » Tue Jan 07, 2014 6:50 pm
Statement 2:
k/3 leaves remainder 2

that means "k" can be
2 ==> n = 2*9 YES
5 ==> n = 5*12 YES
8 ==> n = 8*15 YES
SUFFICIENT

The problem in your solution is that you assumed k = 1, which is incorrect as per the condition. 1/3 leaves remainder "1" and not "2"
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by GMATGuruNY » Tue Jan 07, 2014 7:58 pm
Captain_of_Industry 1984 wrote: 2. When k is divided by 3, the remainder is 2
This means k is 2 more than a multiple of 3:
k = 3a + 2, where a is a nonnegative integer.
Thus:
k = 2, 5, 8, 11, 14, 17...

Given that n = k(k+7), we get:
If k=2, then n = 2*9.
If k=5, then n = 5*12.
If k=8, then n = 8*15.
In every case, n is divisible by 6.
SUFFICIENT.
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by sanju09 » Tue Jan 07, 2014 11:19 pm
Captain_of_Industry 1984 wrote:Hey guys,

I have a question on this D.S. problem, can someone explain why my solution is incorrect?

If k is a positive integer and n = k(k + 7), is n divisible by 6?

1. K is odd

Test cases:
K=1 n=1(1+7) = 8, is 8/6 NO
K=3 n=3(3+7)=30, is 30/6 YES
INSUFFICIENT

2. When k is divided by 3, the remainder is 2

Test Cases:
K=1 1/3=0 remainder 2, n=1(1+7) = 8, is 8/6 NO
K=5 5/3 =1 remainder 2, n=5(5+7) =60, is 60/6 YES
INSUFFICIENT

Combined:
K=1 overlaps - NO
K=5 overlaps - YES

OA - B

I've seen a complex math approach that seems to prove that B is sufficient but why would the test cases produce a different result? What am I missing here?

Thanks in advance.
2. When k is divided by 3, the remainder is 2

It means that k = 3p + 2 for any non-negative integer p. When p = 0, k = 2, and the remainder when 3 divides 2 is 2 only. So n = (3p + 2) (3p + 2 + 7) = (3p + 2) (3p + 9) = 3 (p + 3) (3p + 2). Now, if p is even, 3p + 2 is also even, which makes the expression 3 (p + 3) (3p + 2) divisible by 6. And, if p is odd, p + 3 is even, which makes the expression 3 (p + 3) (3p + 2) again divisible by 6. In other words, if k = 3p + 2, then n is always divisible by 6. Hence, sufficient
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