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For positive integer k, is the expression (k + 2)(k^2 + 4k +

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Re: For positive integer k, is the expression (k + 2)(k^2 + 4k +  [#permalink]

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New post 25 Aug 2017, 21:03
One way to look at statement 1 using divisibility rule:

k is divisible by 8
so k is divisible by 4
k+1 :4 gives remainder of 1
k+2 :4 gives remainder of 2
k+3 :4 gives remainder of 3

None of the factor in the product (k+1)(k+2)(k+3) is divisible by 4, therefore the product is not divisible by 4.
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Re: For positive integer k, is the expression (k + 2)(k^2 + 4k +  [#permalink]

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New post 30 Dec 2017, 06:11
[quote="Bunuel"]


- Doubt

Since k is +ve => min value of k+1 = 2, hence series can be (2,3,4), (3,4,5), (4,5,6) => product is always divisible by 4 ?
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Re: For positive integer k, is the expression (k + 2)(k^2 + 4k +  [#permalink]

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New post 30 Dec 2017, 06:16
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For positive integer k, is the expression (k + 2)(k^2 + 4k +  [#permalink]

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New post 04 May 2018, 04:31
Bunuel niks18 gmatbusters amanvermagmat chetan2u VeritasPrepKarishma


Quote:
For positive integer k, is the expression (k + 2)(k^2 + 4k + 3) divisible by 4?

\((k + 2)(k^2 + 4k + 3)=(k+1)(k+2)(k+3)\), so the expression is the product of three consecutive integers.


(1) k is divisible by 8 --> \(k=8n=even\) --> \((k+1)(k+2)(k+3)=odd*even*odd\). Now, \(k+2=8n+2\), though even, is not a multiple of 4 (it's 2 greater than a multiple of 8), therefore the expression is not divisible by 4. Sufficient.

(2) (k + 1)/3 is an odd integer --> \(k+1=3*odd=odd\) --> \(k=even\) --> \((k+1)(k+2)(k+3)=odd*even*odd\). Now, \(k+2=even\) may or may not be divisible by 8, therefore the expression may or may not be divisible by 8. For example, consider \(k=2\) and \(k=6\). Not sufficient.

Answer: A.


I understood that you used basic properties of even and odd, but did you also use divisibility rules for 8 while analyzing St 1 and
did you plug numbers for st (2)
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For positive integer k, is the expression (k + 2)(k^2 + 4k +  [#permalink]

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New post 04 May 2018, 05:25
adkikani wrote:
Bunuel niks18 gmatbusters amanvermagmat chetan2u VeritasPrepKarishma


Quote:
For positive integer k, is the expression (k + 2)(k^2 + 4k + 3) divisible by 4?

\((k + 2)(k^2 + 4k + 3)=(k+1)(k+2)(k+3)\), so the expression is the product of three consecutive integers.


(1) k is divisible by 8 --> \(k=8n=even\) --> \((k+1)(k+2)(k+3)=odd*even*odd\). Now, \(k+2=8n+2\), though even, is not a multiple of 4 (it's 2 greater than a multiple of 8), therefore the expression is not divisible by 4. Sufficient.

(2) (k + 1)/3 is an odd integer --> \(k+1=3*odd=odd\) --> \(k=even\) --> \((k+1)(k+2)(k+3)=odd*even*odd\). Now, \(k+2=even\) may or may not be divisible by 8, therefore the expression may or may not be divisible by 8. For example, consider \(k=2\) and \(k=6\). Not sufficient.

Answer: A.


I understood that you used basic properties of even and odd, but did you also use divisibility rules for 8 while analyzing St 1 and
did you plug numbers for st (2)


Hello

In a sense, yes, Bunuel has used divisibility rules of 8. If a number is divisible by 8, and you add '2' to it, you get a number of the form 8K + 2, which, though even, will not be a multiple of 4.

For statement 2, Bunuel has already taken examples -highlighted part
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For positive integer k, is the expression (k + 2)(k^2 + 4k +  [#permalink]

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New post 04 May 2018, 06:10
Hi
St1) If by "Divisibility rule of 8" you mean

"If the last three digits of a whole number are divisible by 8, then the entire number is divisible by 8."

.
NO, this rule has not been used. In fact it is not required.
This is explained beautifully by bunuel.

in St 2) plugging different even no, it makes easier to find out as done by bunuel.

adkikani wrote:
Quote:
For positive integer k, is the expression (k + 2)(k^2 + 4k + 3) divisible by 4?

\((k + 2)(k^2 + 4k + 3)=(k+1)(k+2)(k+3)\), so the expression is the product of three consecutive integers.


(1) k is divisible by 8 --> \(k=8n=even\) --> \((k+1)(k+2)(k+3)=odd*even*odd\). Now, \(k+2=8n+2\), though even, is not a multiple of 4 (it's 2 greater than a multiple of 8), therefore the expression is not divisible by 4. Sufficient.

(2) (k + 1)/3 is an odd integer --> \(k+1=3*odd=odd\) --> \(k=even\) --> \((k+1)(k+2)(k+3)=odd*even*odd\). Now, \(k+2=even\) may or may not be divisible by 8, therefore the expression may or may not be divisible by 8. For example, consider \(k=2\) and \(k=6\). Not sufficient.

Answer: A.


I understood that you used basic properties of even and odd, but did you also use divisibility rules for 8 while analyzing St 1 and
did you plug numbers for st (2)

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Re: For positive integer k, is the expression (k + 2)(k^2 + 4k +  [#permalink]

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New post 02 Sep 2018, 17:52
I think this ia high quality question and I agree with the explanation.
Thanks!
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For positive integer k, is the expression (k + 2)(k^2 + 4k +  [#permalink]

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New post 02 Sep 2018, 21:11
(k+1)(k+2)(k+3) , so the expression is the product of three consecutive integers.
hence, it is a multiple of 3! or 6.

product of n consecutive integers is a multiple of n!



This is an important property to be remembered.
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