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

(1) k is odd. If \(k = 1\), then \(n = k(k + 7) = 8\) and n is NOT divisible by 6 but if \(k = 3\), then \(n = k(k + 7) = 30\) and n IS divisible by 6. Not sufficient.

(2) When k is divided by 3, the remainder is 2 --> \(k = 3x + 2\) --> \(n = k(k + 7) = (3x + 2)(3x + 9)=9x^2+33 x+18=3(3x^2+11x)+18\). Notice that \(3x^2+11x\) is even no matter whether x is even or odd, thus \(n=3(3x^2+11x)+18=3*even+(a \ multiple \ of \ 6)=(a \ multiple \ of \ 6)+(a \ multiple \ of \ 6)=(a \ multiple \ of \ 6)\). Sufficient.

Answer: B.

Hope it's clear.
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Here is my solution-
If n=K(K+7), then-

1) k is odd => odd * (odd + Even) = Odd * Odd. We can not say it is multiple of 6 or not. Insufficient.
2) k = 3m +2. So, n=K(K+7) => n= (3m+2)(3m+9)= 3(3m+2)(m+3) => multiple of 3.
If m is odd, m+3 is even. Hence , multiple of 2. Also it is a multiple of 3 => multiple of 6
If m is even, 3m+2 is even. Hence , multiple of 2. Also it is a multiple of 3 => multiple of 6
So B is sufficient.

Given, n= k(k+7K) = 8k^2
now for n to be divisible by 6, k should be divisible by 3.

1. K is odd.
clearly insufficient, for k=1 answer is No.
for k=3, answer is yes.

2. When k is divided by 3, the remainder is 2.
Remainder is 2, so K can never be divisible by 3,
Hence n will not be divisible by 6. So Sufficient.

IMO, B

What about 5? It is a positive integer, when divided by 3 the remainder is 2 and 5(5+7)= 5(12) = 60, divisible by 6.

Hi,
I'm trying to understand this question too. The question I saw had n=K(K+7), not k+ 7K as written in the question above. Can anyone explain this q with the change please?

Thanks!

Hey guys,

I still have a question on this 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

I see the math approach in the posts above but why would the test cases produce a different result? What am I missing here?

Thanks in advance.

1 divided by 3 yields the remainder of 1, not 2: 1=0*3+1.

Does this make sense?
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I still don't see why Statement 2 is Sufficient. If you use 5, the outcome is 60 (divisible by 60), and if you use 8 the outcome is 320 (not divisible by 6). Please explain.

If k=8, then n=k(k+7)=8*15=120, not 320 and 120 is divisible by 6.

The reason why the second statement is sufficient is given here: if-k-is-a-positive-integer-and-n-k-k-7k-is-n-divisible-162594.html#p1290699 Does it make sense?
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1. K is odd

It's clearly insufficient. Test k=1 and k=3.

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

n=k(k+7)=k(k+1+6) --> n=k(k+1)+6k
6k is always divisible by 6. Need to show that k(k+1) is also divisible by 6.

When k is divided by 3, the remainder is 2, then k=2,5,8,11,...

for k=2,5,8,11,..., k(k+1) is always divisible by 6:
k=2: k(k+1)=2*3=6
k=5: 5*6
k=8: 8*9
k=11: 11*12

k(k+1) and 6k are both divisible by 6, therefore n=k(k+1)+6k divisible by 6. sufficient

The answer is B.

Given: n = k(k + 7)
Question: Is n divisible by 6?

(1) k is odd.
If k = 1, n = 8 - Not divisible by 6
If k = 6, n is divisible by 6
Not sufficient

(2) When k is divided by 3, the remainder is 2.
k = (3b+2)
n = (3b+2)(3b+2 + 7) = (3b + 2)(3b + 9) = 3*(3b + 2)(b + 3)
For n to be divisible by 6, it must be divisible by both 2 and 3. We see that it is divisible by 3. Let's see if it is divisible by 2 too i.e. if it is even.
b can be odd or even in this expression. If it is odd, (b+3) will become even because (Odd + Odd = Even). If it is even, (3b+2) will become even because (Even + Even = Even). So in either case, n will be even. So n will be divisible by 3 as well as 2 i.e. it will be divisible by 6.
Sufficient alone.

Answer (B)
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Bunuel,

Can you explain why : \(3x^2+11x\) is even no matter whether x is even or odd? I'm sure there is a simple theoretical way to see this quicker than plugging in numbers.

Thanks!

\(3x^2+11x=x(3x+11)\).

If x is even the result is obviously even: \(x(3x+11)=even*integer=even\);
If x is odd, then \(x(3x+11)=odd(odd*odd+odd)=odd*even=even\).

Hope it's clear.
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

(1) k is odd.

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

There are 2 variables (n,k) and 1 equation (n=k(k+7)) in the original condition, 2 more equations in the given conditions, so there is high chance (D) will be our answer.
For condition 1, the answer is 'no' for k=1, but 'yes' for k=5, so this is insufficient.
For condition 2, k=3m+2 (m is a positive integer), or k+7=3m+2+7=3(m+3)=a multiple of 3 and k(k+7). This is always even and a multiple of 3, which means its always divisible by 6. This answers the question 'yes'; this is sufficient, and the answer is B.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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VeritasPrepKarishma

Hello Karishma! A question here.. how do you quickly pull the 3 out of the initial statement?
(3b + 2)(3b + 9) = this one-> 3*(3b + 2)(b + 3)

Thank you!

Both terms of (3b + 9) have 3 as a factor.
(3*b + 3*3)

So you pull out a 3 from the brackets.
3*(b + 3)

Initial expression becomes
(3b + 2) * 3 * (b + 3)

The three terms are multiplied so you can arrange them in any way you like.

3 * (3b + 2) * (b + 3)
or
(b + 3) * (3b + 2) * 3
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Statement 1:
Case 1: k=1, with the result that n=1(1+7) = 8
In this case, n is not divisible by 6, so the answer to the question stem is NO.
Case 2: k=2, with the result that n = 2(2+7) = 18
In this case, n is divisible by 6, so the answer to the question stem is YES.
Since the answer is NO in Case 1 but YES in Case 2, INSUFFICIENT.

Statement 2:
k = 3x+2, where x is an nonnegative integer, with the result that n = (3x+2)(3x+9) = (3x+2)(3)(x+3)
The resulting expression in blue has a factor of 3 and thus is a MULTIPLE OF 3.
If x is EVEN, then 3x+2 = EVEN.
If x is ODD, then x+3 = EVEN.
Either way, the blue expression is an EVEN MULTIPLE OF 3, with the result that n is divisible by 6.
Thus, the answer to the question stem is YES.
SUFFICIENT.


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Bunuel, GMAT Prep question. Please tag it