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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.

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+ 7K as written in the question above. Can anyone explain this q with the change please?

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]

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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.

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.

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]

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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

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

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]

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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.

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.

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]

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11 Nov 2015, 08:30

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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 [Sufficient] <--- Answer is B

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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