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

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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

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?

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?

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)

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.

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

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.

Hi Bunuel,

from this "2. When k is divided by 3, the remainder is 2"
Then K can be 2,5,8,11,.... and when we pluged in these numbers to K(K+7)/6, it will have both devisable and non-divisible.
Why its insufficient?

Yes, from (2) k can be: 2, 5, 8, 11, ... But:

If k = 2, then k(k + 7) = 18, which IS divisible by 6;
If k = 5, then k(k + 7) = 60, which IS divisible by 6;
If k = 8, then k(k + 7) = 120, which IS divisible by 6;
If k = 11, then k(k + 7) = 11*18, which IS divisible by 6;
...

For every possible value of k, k(k + 7) will be divisible by 6.

"Also it is a multiple of 3." What is it?