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If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]
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02 Nov 2013, 14:01
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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.
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Last edited by AbhiJ on 10 Apr 2014, 06:42, edited 2 times in total.
Renamed the topic, edited the question and added the OA.



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Re: If k is a positive integer and n = k(k+7)..... [#permalink]
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10 Nov 2013, 12:30
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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. 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+ 7 K 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. Answer: B. 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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11 Nov 2013, 03:23
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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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04 Nov 2013, 17:51
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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 + 7k), is n divisible [#permalink]
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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. Answer (B)
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If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]
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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 [Sufficient] < Answer is B



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Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]
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29 Mar 2016, 22:34
iliavko wrote: VeritasPrepKarishmaHello 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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Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]
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30 Mar 2016, 08:08
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Thank you a lot for the reply! The situation is much simpler than I thought then lol
Regards!



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Re: If k is a positive integer and n = k(k+7)..... [#permalink]
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02 Nov 2013, 22:08
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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. 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



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Re: If k is a positive integer and n = k(k+7)..... [#permalink]
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10 Nov 2013, 08:45
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+ 7 K 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, 15:41
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.
Answer: B.
Hope it's clear. Thanks, that helps!



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Re: If k is a positive integer and n = k(k+7)..... [#permalink]
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07 Jan 2014, 15:01
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?
Thanks in advance.



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Re: If k is a positive integer and n = k(k+7)..... [#permalink]
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08 Jan 2014, 03:49
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?
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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Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]
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09 Apr 2014, 16:51
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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Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink]
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10 Apr 2014, 02:58
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, 19:16
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
The answer is B.



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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, 19:38
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 answerHi 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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05 May 2015, 11:03
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.
Answer is B



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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, 22:22
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. Thanks!




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