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 500,000 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: If k is an integer, is 2^k + 3^k = m ? [#permalink]

Show Tags

30 Aug 2013, 01:26

danzig wrote:

If k is an integer, is 2^k + 3^k = m ?

(1) 4^k + 9^k = m^2 - 12

(2) k = 1

I don't agree with the OA. If we use statements 1 and 2 together, we get this:

We replace the value of \(k\) in the original question, so the question now is:\(is k = 5 ?\)

Now we replace the value of \(k\) in statement (1), so:

\(m^2 = 25\)

So,\(m = +/- 25\)

There is not indication whether m is possitive.

IMO, the answer: E

I don't know what was the original OA but I am assuming it was C, here's how I think the answer should be C, please do correct if there is a flaw in my reasoning.

Given If k is an integer, is \(2^k + 3^k = m\)? Squaring both sides original equation now becomes \(2^{2k} +3^{2k}+2.6^k=m^2\)

statement 1:\(4^k + 9^k = m^2 - 12\)this can be written as \(2^{2k} +3^{2k}+12 =m^2\)..

Comparing this with original equation we see that this will be equal to original equation only if K=1 since we do not have the value of K, hence insufficient

statement 2 : K=1 , by itself it is insufficient as we do not know the value of M

1+ 2

K=1 then \(m^2\) = 25

this is also what we get from the original equation,doesn't matter what m is, \(m^2\) is 25 Please share your views _________________

Re: If k is an integer, is 2^k + 3^k = m ? [#permalink]

Show Tags

30 Aug 2013, 04:41

1

This post received KUDOS

Expert's post

stne wrote:

danzig wrote:

If k is an integer, is 2^k + 3^k = m ?

(1) 4^k + 9^k = m^2 - 12

(2) k = 1

I don't agree with the OA. If we use statements 1 and 2 together, we get this:

We replace the value of \(k\) in the original question, so the question now is:\(is k = 5 ?\)

Now we replace the value of \(k\) in statement (1), so:

\(m^2 = 25\)

So,\(m = +/- 25\)

There is not indication whether m is possitive.

IMO, the answer: E

I don't know what was the original OA but I am assuming it was C, here's how I think the answer should be C, please do correct if there is a flaw in my reasoning.

Given If k is an integer, is \(2^k + 3^k = m\)? Squaring both sides original equation now becomes \(2^{2k} +3^{2k}+2.6^k=m^2\)

statement 1:\(4^k + 9^k = m^2 - 12\)this can be written as \(2^{2k} +3^{2k}+12 =m^2\)..

Comparing this with original equation we see that this will be equal to original equation only if K=1 since we do not have the value of K, hence insufficient

statement 2 : K=1 , by itself it is insufficient as we do not know the value of M

1+ 2

K=1 then \(m^2\) = 25

this is also what we get from the original equation,doesn't matter what m is, \(m^2\) is 25 Please share your views

It does matter. If m=-5, then the question is: does \(2^k + 3^k = -5\)? And you cannot square this. _________________

Re: If k is an integer, is 2^k + 3^k = m ? [#permalink]

Show Tags

30 Aug 2013, 07:00

Bunuel wrote:

stne wrote:

danzig wrote:

If k is an integer, is 2^k + 3^k = m ?

(1) 4^k + 9^k = m^2 - 12

(2) k = 1

I don't agree with the OA. If we use statements 1 and 2 together, we get this:

We replace the value of \(k\) in the original question, so the question now is:\(is k = 5 ?\)

Now we replace the value of \(k\) in statement (1), so:

\(m^2 = 25\)

So,\(m = +/- 25\)

There is not indication whether m is possitive.

IMO, the answer: E

I don't know what was the original OA but I am assuming it was C, here's how I think the answer should be C, please do correct if there is a flaw in my reasoning.

Given If k is an integer, is \(2^k + 3^k = m\)? Squaring both sides original equation now becomes \(2^{2k} +3^{2k}+2.6^k=m^2\)

statement 1:\(4^k + 9^k = m^2 - 12\)this can be written as \(2^{2k} +3^{2k}+12 =m^2\)..

Comparing this with original equation we see that this will be equal to original equation only if K=1 since we do not have the value of K, hence insufficient

statement 2 : K=1 , by itself it is insufficient as we do not know the value of M

1+ 2

K=1 then \(m^2\) = 25

this is also what we get from the original equation,doesn't matter what m is, \(m^2\) is 25 Please share your views

It does matter. If m=-5, then the question is: does \(2^k + 3^k = -5\)? And you cannot square this.

Well if you say so, there are many ways algebraic equations can be manipulated and I thought they could be squared .Thank you +1 _________________

Re: If k is an integer, is 2^k + 3^k = m ? [#permalink]

Show Tags

25 Dec 2014, 07:40

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

HBS alum talks about effective altruism and founding and ultimately closing MBAs Across America at TED: Casey Gerald speaks at TED2016 – Dream, February 15-19, 2016, Vancouver Convention Center...

By Libby Koerbel Engaging a room of more than 100 people for two straight hours is no easy task, but the Women’s Business Association (WBA), Professor Victoria Medvec...