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

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[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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Tanvi94 wrote:
Bunuel wrote:

- 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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(5, 6, 7) is not.
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For positive integer k, is the expression (k + 2)(k^2 + 4k +  [#permalink]

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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.

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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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.

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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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.

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.

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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GMAT 1: 750 Q50 V41 Re: For positive integer k, is the expression (k + 2)(k^2 + 4k +  [#permalink]

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

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_________________ Re: For positive integer k, is the expression (k + 2)(k^2 + 4k +   [#permalink] 05 Sep 2019, 23:18

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