If k is a positive integer and n = k(k + 7k), is n divisible : GMAT Data Sufficiency (DS)
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# If k is a positive integer and n = k(k + 7k), is n divisible

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02 Nov 2013, 13:01
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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.
[Reveal] Spoiler: OA

Last edited by AbhiJ on 10 Apr 2014, 05:42, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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02 Nov 2013, 21:08
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Jem2905 wrote:
Hi guys, trying to get a little help on this problem that stumped me recently on a practice test. After going back and spending some more time with it, I got a different answer but I'm not sure if it's the right answer, and I'm not exactly sure I understand why it's the correct answer... any resphrasing of the question or statements will be hugely appreciated. Thanks!!

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

(1) k is odd.

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

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.

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
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04 Nov 2013, 16:51
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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.
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Re: If k is a positive integer and n = k(k+7)..... [#permalink]

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10 Nov 2013, 07:45
Chiranjeevee wrote:
Jem2905 wrote:
Hi guys, trying to get a little help on this problem that stumped me recently on a practice test. After going back and spending some more time with it, I got a different answer but I'm not sure if it's the right answer, and I'm not exactly sure I understand why it's the correct answer... any resphrasing of the question or statements will be hugely appreciated. Thanks!!

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

(1) k is odd.

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

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.

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!
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Re: If k is a positive integer and n = k(k+7)..... [#permalink]

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10 Nov 2013, 11:30
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ashsim wrote:
Chiranjeevee wrote:
Jem2905 wrote:
Hi guys, trying to get a little help on this problem that stumped me recently on a practice test. After going back and spending some more time with it, I got a different answer but I'm not sure if it's the right answer, and I'm not exactly sure I understand why it's the correct answer... any resphrasing of the question or statements will be hugely appreciated. Thanks!!

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

(1) k is odd.

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

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.

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!

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.

Hope it's clear.
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Re: If k is a positive integer and n = k(k+7)..... [#permalink]

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10 Nov 2013, 14:41
Quote:
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.

Hope it's clear.

Thanks, that helps!
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11 Nov 2013, 02:23
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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.
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Re: If k is a positive integer and n = k(k+7)..... [#permalink]

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07 Jan 2014, 14:01
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?

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Re: If k is a positive integer and n = k(k+7)..... [#permalink]

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08 Jan 2014, 02:49
msbandi4321 wrote:
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?

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

Does this make sense?
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Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]

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09 Apr 2014, 15:51
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.
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Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]

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10 Apr 2014, 01:22
jbartuccio wrote:
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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Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]

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10 Apr 2014, 01:58
jbartuccio wrote:
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 you refer to the initial question:
n= k(k+7k) = 8k^2

Hence, if k = 5, n = 8*5*5
if k = 8, n = 8*8*8

both of them are clearly not divisible by 6.

*press kudos if you like the answer
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Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]

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02 Feb 2015, 18:16
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

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Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]

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02 Feb 2015, 18:38
ricsingh wrote:
jbartuccio wrote:
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 you refer to the initial question:
n= k(k+7k) = 8k^2

Hence, if k = 5, n = 8*5*5
if k = 8, n = 8*8*8

both of them are clearly not divisible by 6.

*press kudos if you like the answer

Hi ricsingh:

n=k(k+7) is not same as n=k(k+7k).

Watch out for silly mistakes!

Posted from my mobile device
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Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]

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02 Feb 2015, 21:29
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Jem2905 wrote:
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.

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.

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Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]

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05 May 2015, 10:03
n= k(k+7) so is n divisible by 6?

State 1: K is Odd
Possible values are 1, 3, 5, 7 etc
Expression is not true if n = 1 but true for rest of the numbers so Insufficient

State 2: When K is divided by 3, the remainder is 2.
Possible values are 2, 8, 11, 14, 17 etc
For all there values k(k+7) is divisible by 6 - Hence State 2 is Sufficient.

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Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]

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10 Nov 2015, 21:22
Bunuel wrote:

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.

Hope it's clear.

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!
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Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]

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10 Nov 2015, 22:35
dubyap wrote:
Bunuel wrote:

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.

Hope it's clear.

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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Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]

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10 Nov 2015, 22:51
Bunuel wrote:
dubyap wrote:
Bunuel wrote:

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.

Hope it's clear.

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.

Very clear. Thank you!
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If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]

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11 Nov 2015, 07:30
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Jem2905 wrote:
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.

Divisibility by 6 means must be divisible by 3x2 (Prime Factors of 6)

1) k is Odd
==> k(k+7) = Odd(Odd+Odd) = Even
This will give me at least one factor of 2.. but nothing about factor of 3

[Insufficient]

2) k Divided by 3 ==> Remainder = 2
Now Possible values could be
2,5,8,11 -- Alternatively Even and Odd
So if k=2,8,...
Then I am getting One factor of 2 and when I add these to 7 I get Multiple of 3,

Similarly when I I choose k=5,11.. Odd Number with remainder of 2 and I add them to 7 I get Even Multiple of 3 i.e. Multiple of 6

This is happening because, Numbers with Remainders of 2 when added to 7, which leaves remainder of 1 when divided by 3, makes the Sum Divisible by 3
If k is a positive integer and n = k(k + 7k), is n divisible   [#permalink] 11 Nov 2015, 07:30

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