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Re: If k is a positive integer and n = k(k+7)..... [#permalink]
02 Nov 2013, 21:08
2
This post was BOOKMARKED
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. 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.
Re: If k is a positive integer and n = k(k+7)..... [#permalink]
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. 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?
Re: If k is a positive integer and n = k(k+7)..... [#permalink]
10 Nov 2013, 11:30
10
This post received KUDOS
Expert's post
3
This post was BOOKMARKED
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. 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!
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.
Re: If k is a positive integer and n = k(k+7)..... [#permalink]
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.
Re: If k is a positive integer and n = k(k+7)..... [#permalink]
11 Nov 2013, 02:23
4
This post received KUDOS
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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.
Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]
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.
Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]
10 Apr 2014, 01:22
Expert's post
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.
Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]
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
Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]
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
Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]
02 Feb 2015, 21:29
Expert's post
1
This post was BOOKMARKED
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.
Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]
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.
Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]
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.
Answer: B.
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.
Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]
10 Nov 2015, 22:35
Expert's post
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.
Answer: B.
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 \(x(3x+11)=odd(odd*odd+odd)=odd*even=even\).
Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]
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.
Answer: B.
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 \(x(3x+11)=odd(odd*odd+odd)=odd*even=even\).
If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]
11 Nov 2015, 07:30
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 [Sufficient] <--- Answer is B
gmatclubot
If k is a positive integer and n = k(k + 7k), is n divisible
[#permalink]
11 Nov 2015, 07:30
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