Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

3

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.

Re: integer k ... DS [#permalink]
22 Sep 2009, 16:03

I agree that the correct answer is A.

Is (k + 2)(k2 + 4k + 3) divisible by 4?

(k + 2)(k^2 + 4k + 3) = (k + 1)(k + 2)(k + 3)

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

Subbing this into the original equation:

(k+1)(k+2)(k+3) = (6n + 3)(6n + 4)(6n + 5)

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.

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

1

This post received KUDOS

Expert's post

1

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 + 2)(k^2 + 4k + [#permalink]
06 Nov 2014, 15:03

Hi,

I had question when it came to the prime factors of (K+2). If k is divisible by 8, that means that it's divisible by 4, in every single instance. Correct? If it's divisible by 4, then it has all the PF's of 2, in ever single instance. Correct? If a sum is a multiple of a 2 plus a multiple of 2, then doesn't that mean that the sum is a multiple of 4?

Re: For positive integer k, is the expression (k + 2)(k^2 + 4k + [#permalink]
07 Nov 2014, 03:37

Expert's post

russ9 wrote:

Hi,

I had question when it came to the prime factors of (K+2). If k is divisible by 8, that means that it's divisible by 4, in every single instance. Correct? If it's divisible by 4, then it has all the PF's of 2, in ever single instance. Correct? If a sum is a multiple of a 2 plus a multiple of 2, then doesn't that mean that the sum is a multiple of 4?

Sorry, but your question is not clear at all... _________________

My three goals of business school: entrepreneurship, network, and professor mentor. I want to build something. I want to meet new people and create life-long friendships. I want to...