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:

If \(p\) is a prime number, what is the value of \(p\)?

(1) \(\sqrt[3]{1-p^2}=-n\), where \(n\) is a prime number. (2) \(8p^2+1=m\), where \(m\) is a prime number.

Please find the solution as attached.

Answer: option D

Attachments

File comment: www.GMATinsight.com

Untitled2.jpg [ 137.45 KiB | Viewed 768 times ]

_________________

Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html

Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]

Show Tags

03 Oct 2010, 12:06

1

This post received KUDOS

A is ans on solving a we get 1+ n^3 = P^2 so it will be satisfied with n=2 and p=3.. the reason is..p will be alws greater than n..so p will alws be odd no..so its square will be alws odd... so n^3 must be even which is possible with only prime no 2.. so 1+n^3 will bcome odd... so satisfied...

2) on solving eq bcom 8p^2 +1 =m..so I m unable to find any prime no which is 24k+1..where k is any + integer.. so not satisfied

Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]

Show Tags

03 Oct 2010, 12:11

1

This post received KUDOS

Bunuel wrote:

If \(p\) is a prime number, what is the value of \(p\)?

(1) \(\sqrt[3]{1-p^2}=-n\), where \(n\) is a prime number. (2) \(8p^2+1=m\), where \(m\) is a prime number.

(1) \(\sqrt[3]{1-p^2}=-n\) OR \(p^2=1+n^3=(1+n)(n^2-n+1)\) If n=2, p=3 If n>2, then n is odd, (n+1) is even, hence p^2 is even, but this is only possible if p=2 and if p=2, n is cube_root(-3) which is not an integer So only possibility is p=3 Sufficient

(2) \(8p^2+1=m\) p is an integer and a prime so it can either be of the form 3k+1 or 3k-1 or be 3 (all other numbers of form 3k are composite) 8*(3k+1)^2+1=8*(9k^2+6k+1)+1=3*(24k^2+16k+3) .. which cannot be a prime since k>=1 8*(3k-1)^2+1=8*(9k^2-6k+1)+1=3*(24k^2-16k+3) .. which cannot be a prime since k>=1 8*3^2+1=73 is a prime So only possible value of p is 3 Sufficient

Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]

Show Tags

03 Oct 2010, 12:29

sudhanshushankerjha wrote:

hey Man p is an integer and a prime so it can either be of the form 6k+1 or 6k-1... not 3k+1 and 3k-1

You can always divide all integers into 3 sets : 3k , 3k+1 , 3k-1 In case of primes, it is easy to see only one element in the first set, i.e., 3
_________________

If \(p\) is a prime number, what is the value of \(p\)?

(1) \(\sqrt[3]{1-p^2}=-n\), where \(n\) is a prime number. (2) \(8p^2+1=m\), where \(m\) is a prime number.

I wrote this question, so below is my solution:

(1) \(\sqrt[3]{1-p^2}=-n\), where \(n\) is a prime number.

It's clear that \(p\neq{2}\) as in this case \(\sqrt[3]{1-p^2}\neq{integer}\), so \(p\) is a prime more than 2, so odd. Then \(1-p^2=odd-odd=even\) --> cube root from even is either an even integer or not an integer at all, we are told that \(\sqrt[3]{1-p^2}=-n=integer\), so \(-n=even=-prime\) --> \(n=2\) (the only even prime) --> \(\sqrt[3]{1-p^2}=-2\) --> \(p=3\). Sufficient.

(2) \(8p^2+1=m\), where \(m\) is a prime number.

\(p\) can not be any prime but 3 for \(8p^2+1\) to be a prime number: because if \(p\) is not 3, then it's some other prime not divisible by 3, but in this case \(p^2\) (square of a not multiple of 3) will yield the remainder of 1 when divided by 3 (not a multiple of 3 when squared yields remainder of 1 when divided by 3), so \(p^2\) can be expressed as \(3k+1\) --> \(8p^2+1=8(3k+1)+1=24k+9=3(8k+3)\) --> so \(8p^2+1\) becomes multiple of 3, so it can not be a prime. So \(p=3\) and in this case \(8p^2+1=73=m=prime\). Sufficient.

Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]

Show Tags

05 Oct 2010, 10:55

I am really confused. Can someone please explain me,

Statement 2:

I understand that if we work out the 8P^2 + 1 : where P = 2 we get m = 33, which is NOT a prime.

Great! I am on-board till here.

Is the strategy to try every prime number from 2, onwards??

How is the value of P chosen??

You state P cannot be any prime but 3 for 8P^2 + 1 to be a prime number. How can you be so sure? I do not have an exhaustive list of all prime number, but this claim is little far fetched.

I am really confused. Can someone please explain me,

Statement 2:

I understand that if we work out the 8P^2 + 1 : where P = 2 we get m = 33, which is NOT a prime.

Great! I am on-board till here.

Is the strategy to try every prime number from 2, onwards??

How is the value of P chosen??

You state P cannot be any prime but 3 for 8P^2 + 1 to be a prime number. How can you be so sure? I do not have an exhaustive list of all prime number, but this claim is little far fetched.

Please clear my confusion.

There are infinitely many prime numbers so there is no "list of all primes".

Next, \(p\) can not be any prime but 3 for \(8p^2+1\) to be a prime number: because if \(p\) is not 3, then it's some other prime not divisible by 3, but in this case \(p^2\) (square of a not multiple of 3) will yield the remainder of 1 when divided by 3 (not a multiple of 3 when squared yields remainder of 1 when divided by 3), so \(p^2\) can be expressed as \(3k+1\) --> \(8p^2+1=8(3k+1)+1=24k+9=3(8k+3)\) --> so \(8p^2+1\) becomes multiple of 3, so it can not be a prime. So \(p=3\) and in this case \(8p^2+1=73=m=prime\).
_________________

Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]

Show Tags

05 Oct 2010, 12:47

Hi Bunuel,

OK. But I am still not clear about why is it important to be divisible by 3.

"because if \(p\) is not 3, then it's some other prime not divisible by 3, but in this case \(p^2\) (square of a not multiple of 3) will yield the remainder of 1 when divided by 3 (not a multiple of 3 when squared yields remainder of 1 when divided by 3)"

Bunuel wrote:

There are infinitely many prime numbers so there is no "list of all primes".

Next, \(p\) can not be any prime but 3 for \(8p^2+1\) to be a prime number: because if \(p\) is not 3, then it's some other prime not divisible by 3, but in this case \(p^2\) (square of a not multiple of 3) will yield the remainder of 1 when divided by 3 (not a multiple of 3 when squared yields remainder of 1 when divided by 3), so \(p^2\) can be expressed as \(3k+1\) --> \(8p^2+1=8(3k+1)+1=24k+9=3(8k+3)\) --> so \(8p^2+1\) becomes multiple of 3, so it can not be a prime. So \(p=3\) and in this case \(8p^2+1=73=m=prime\).

I guess, you will have to explain this a little more.
_________________

OK. But I am still not clear about why is it important to be divisible by 3.

"because if \(p\) is not 3, then it's some other prime not divisible by 3, but in this case \(p^2\) (square of a not multiple of 3) will yield the remainder of 1 when divided by 3 (not a multiple of 3 when squared yields remainder of 1 when divided by 3)"

I guess, you will have to explain this a little more.

Because if \(p\) IS NOT divisible by 3 then \(8p^2+1\) IS divisible by 3 and thus can not be a prime, so \(p\) must be divisible by 3, only prime divisible by 3 is 3.
_________________

Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]

Show Tags

23 Oct 2010, 19:32

Tough one, I could solve the question by plugging in the numbers but didn't think of trying (especially for 2nd statement) the above concepts, despite knowing for a number to be prime it should be either \(6k+1\) or \(6k-1\). Similarly, for statemet1 despite knowing that \(p\) can not be \(2\) and it would be more than didn't think that it must be odd and apply Odd & Even concept here.

Thank you! for reminding me to use concepts rather than just guessing and then getting doubtful solutions (and doing silly mistakes).
_________________

"I choose to rise after every fall" Target=770 http://challengemba.blogspot.com Kudos??

Re: If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]

Show Tags

14 Jan 2015, 00:54

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.
_________________

If p is a prime number, what is the value of p ? (1) \sqrt [#permalink]

Show Tags

21 Jul 2016, 09:16

GMATinsight wrote:

Bunuel wrote:

If \(p\) is a prime number, what is the value of \(p\)?

(1) \(\sqrt[3]{1-p^2}=-n\), where \(n\) is a prime number. (2) \(8p^2+1=m\), where \(m\) is a prime number.

Please find the solution as attached.

Answer: option D

Dear GMATinsight,

Thanks for you explanation. I did the same as you did in solving those question. But I had a doubt that there maybe a higher prime number that satisfy fact 2 so I kept doing this to 13 & 17. How should I be confident to stop at 11 like you?

If \(p\) is a prime number, what is the value of \(p\)?

(1) \(\sqrt[3]{1-p^2}=-n\), where \(n\) is a prime number. (2) \(8p^2+1=m\), where \(m\) is a prime number.

Please find the solution as attached.

Answer: option D

Dear GMATinsight,

Thanks for you explanation. I did the same as you did in solving those question. But I had a doubt that there maybe a higher prime number that satisfy fact 2 so I kept doing this to 13 & 17. How should I be confident to stop at 11 like you?

1) You need to find out a pattern in solutions because there are always patterns followed among all acceptable or unacceptable solutions

2) To find a pattern you need three instances/examples to observe

I saw the pattern in solutions that each of the solutions that I was finding was a multiple of 3 so I concluded that this pattern is going to continue

Trusting the pattern based on three instances usually gets you the correct answer in 99% cases and the remaining 1% are usually made by Manhattan intentionally to prove this thought process incorrect. Thankfully Manhattan doesn't make GMAT so the chances are bleak that you get wrong answer trusting the pattern
_________________

Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html

Version 8.1 of the WordPress for Android app is now available, with some great enhancements to publishing: background media uploading. Adding images to a post or page? Now...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...