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Re: integer k ... DS [#permalink]
22 Sep 2009, 07:31

4

This post received KUDOS

Start off by factoring the original equation in the stem where (k^2+4k+3)=(k+3)(k+1) So, (k+2(k^2+4k+3)=(k+1)(k+2)(k+3) where (k+1)(k+2)(k+3) are essentially consecutive integers.

As consecutive integers, (k+1)(k+2)(k+3) can be (odd)(even)(odd) in which case k has to be even. It can also be (even)(odd)(even) in which case k has to be odd. If k is odd,(k+1)(k+2)(k+3) will definitely be divisible by 4 as the product terms carry at least two 2s. This is because there are two even numbers in the product terms and they will each carry at least one 2 in their prime factorization and 4=(2)(2).

If k is even however,(k+1)(k+2)(k+3) may or may not be divisible by 4. It depends if the middle term carries two 2s, (i.e. is a multiple of 4) or not.(k+1) and (k+3) will definitely not carry a 2 in this case as they are odd numbers. For example, if k=2,(k+1)(k+2)(k+3) can be (3)(4)(5) which is divisible by 4 as the middle term, 4, is a multiple of 4. The same logic applies for k=6. If k=8 however,(k+1)(k+2)(k+3) = (9)(10)(11) which is NOT divisible by 4 as the middle term is 10=5x2 and there is only one 2 in its prime factorization.

1) All multiples of 8 are even. Hence, k is even. Possible values of (k+1)(k+2)(k+3) are: If k=8, (k+1)(k+2)(k+3) = (9)(10)(11) If k =16, (k+1)(k+2)(k+3) =(17)(18)(19) If k=24,(k+1)(k+2)(k+3) = (25)(26)(27) If k=32, (k+1)(k+2)(k+3) =(33)(34)(25)

Looking at the middle, or the only even term, of each series of consecutive integers, 10, 18 or 26, we see that in their prime factorization, they only contain one 2. Hence, the entire product of 3 consecutive integers will NOT be divisible by 4 if k is a multiple of 8.

Logically, this is also because multiples of 4 which are 8 or greater include - 8,12,16,20, 24 Multiples of 8 include 8, 16, 24. The equation given in the stem asks for the product of 3 consecutive integers. The difference of multiples of 4 and 8 are either 0 or a multiple of 8 apart. Never 3 apart. Hence, the middle number of (k+1)(k+2)(k+3) when k is a multiple of 8 will never contain a multiple of 4. Statement A is sufficient.

2) I assume the statement means (k+1)/3 is odd. So, k+1=3(odd number) k+1=odd k=odd-1 k = even

For reasons explained in the beginning, if k is even,(k+1)(k+2)(k+3) may or may not be divisible by 4.

Now to simplify matters, one of the terms NEEDS to be divisible by 4 in order for the whole expression to be divisible by 4. Why?

In order for an integer to be divisible by 4, its product must contain two factors of 2. This would appear to be satisfied by two even numbers. In the instance that we had two even numbers (only possible in this case if k+1 and k+3 are even, then we would already have a number divisible by 4 since those numbers are two consecutive even numbers.

Statement 1: k is divisible by 8:

If k is divisible by 8, then k = 8n, where n is a positive integer. This immediately makes k+1 and k+3 odd numbers, which are not divisible by 4. Now we test the (k+2) term:

\(k+2 = 8n + 2 = 4(2n) + 2\)

As a result, if k is divisible by 8, the expression when divided by 4 will have a remainder of 2. Therefore we can can conclude it IS NOT divisible by 4, and Statement 1 is sufficient.

Statement 2: (k + 1)/3 is an odd integer:

We can represent an odd integer as 2n + 1, where n is an integer. Therefore, \((k+1)/3 = 2n + 1\) \(k = 6n + 2\)

Since 6n + 3 and 6n + 5 both represent odd numbers, if 6n + 4 is divisible by 4, the expression is divisible by 4. Since we cannot determine this (if n = 1, it is not divisible by 4, if n = 2, it is divisible by 4, etc.), Statement B is insufficient.

Please explain. A little doubtful with this OA _________________

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Re: For positive integer k, is the expressio [#permalink]
21 Feb 2010, 13:55

Expert's post

1) if k is divisible by 8 (and by 4 too), then k+2 is even but isn't divisible by 4 or 8. (k^2+4k+3) - an odd integer. ---> (even but not divisible by 4 or 8)*odd ---> the expression isn't divisible by 4.

2)k+1/3 is odd --> k+1 is odd --> k is even --> k+2 is even. if k+2: - is divisible by 4, the expression is also divisible by 4 - isn't divisible by 4, the expression isn't divisible by 4 insufficient _________________

Re: For positive integer k, is the expressio [#permalink]
12 Mar 2010, 07:30

IMO A

expression = (k+2)(k+1)(k+3) (1): (k+1)(k+3) is odd integer, (k+2) is only divisible by 2 but not by 4 so (1) is sufficient! (2): k=2 then expression=4*odd integer is divisible by 4 k=5 then expression=6*odd integer is not divisible by 4 _________________

Re: For positive integer k, is the expressio [#permalink]
13 Jul 2010, 03:56

Another way to look at it: i)k is divisible by 8, so k is also divisible by 4. From the property of consecutive multiple , we know that the next term that is divisible by 4 after k is (k+4) (that is every 4th number from k). That means (K+2)(K+1)(K+3) is not divisible by 4.

ii) k+2 is even, but we don't know if it is a multiple of 4 or not. Plug in 1 and 3 for k+1 yields different answer

Re: For positive integer k, is the expression (k + 2)(k2 + 4k + [#permalink]
22 Apr 2014, 02:10

2

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

pretzel wrote:

IMO D is the answer.

For second statement, if K + 1/3 or (K+1)/3, K is even. And so, in (K + 2) (K^2 + 4K + 3) = (K+2)(K+3)(K+1) = even x odd x odd

Am I correct?

No, the correct answer is A.

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.

Re: For positive integer k, is the expression (k + 2)(k2 + 4k + [#permalink]
29 Apr 2014, 15:52

Bunuel wrote:

pretzel wrote:

IMO D is the answer.

For second statement, if K + 1/3 or (K+1)/3, K is even. And so, in (K + 2) (K^2 + 4K + 3) = (K+2)(K+3)(K+1) = even x odd x odd

Am I correct?

No, the correct answer is A.

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 8 (it's 2 greater than a multiple of 8), therefore the expression is not divisible by 8. 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.

Hope it's clear.

I think that we are asked if the product of the three consecutive integers is divisible by 4 not by 8

Re: For positive integer k, is the expression (k + 2)(k2 + 4k + [#permalink]
30 Apr 2014, 06:38

Expert's post

jlgdr wrote:

Bunuel wrote:

pretzel wrote:

IMO D is the answer.

For second statement, if K + 1/3 or (K+1)/3, K is even. And so, in (K + 2) (K^2 + 4K + 3) = (K+2)(K+3)(K+1) = even x odd x odd

Am I correct?

No, the correct answer is A.

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 8 (it's 2 greater than a multiple of 8), therefore the expression is not divisible by 8. 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.

Hope it's clear.

I think that we are asked if the product of the three consecutive integers is divisible by 4 not by 8

Let me know Cheers! J

It should have been 4 instead of 8... _________________

Re: For positive integer k, is the expression (k + 2)(k^2 + 4k + [#permalink]
09 May 2014, 08:31

Hi Bunuel,

In statement 2, if we use k=6 then it does not meet the criteria (k+1)/3= Odd integer. However, if we use k=2 or k=8 or k=20 etc which meet the criteria, then we are able to say that statement 2 is sufficient. Is this line of thinking correct?

Re: For positive integer k, is the expression (k + 2)(k^2 + 4k + [#permalink]
29 Jul 2014, 10:20

(1): k is divisible by 8: so k is even in the expression (k + 2) (k^2 + 4k + 3), k^2 + 4k + 3 is odd if k =8 how? k^2 = 8^2 = even, 4k = even, 3 = odd so even + even + odd = odd and odd is never divisible by 4 it is sufficient if we test only k+2 is divisible by 4 since k is a multiple of 8, k is also a multiple of 4 (factor foundation rule) so a multiple of 4 + 2 is not divisible by 4. Sufficient (2) (k+1) / 3 = odd k + 1 = 3* odd = odd => k = odd - 1 = even if k = 2 the expression is divisible by 4 if k = 4 the expression is not divisble by 4. Not sufficient

Re: For positive integer k, is the expression (k + [#permalink]
31 Jan 2015, 15:54

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