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:

Find the number of pairs of positive integers (x, y) such that x^6 = y^2 + 127.

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

pls share if u used a quick approach to solve this

First note that: x and y are +ve integers i.e. x and y cannot be -ves, zeros and fractions.

Given that:- x^6 = y^2 + 127 x can have a value of 1....................n, however x>2 because:

If x = 1, x^6 = 1 and y = sqrt(1-127). y is an irrational number. Not possible. If x = 2, x^6 = 64 and y = sqrt(64-127). y is an irrational number. Not possible. If x = 3, x^6 = 729 and y = sqrt(729-127) = sqrt(602). Now y is a fraction. Not possible. If x = 4, x^6 = 64x64 and y = sqrt{(64x64) -127} = .......?

Now its not possible to do any calculation beyond this point in 2-5 minuets. Until and unless there is a quick approach, I would say this not a gmat-type question. I would love to see if anybody has a quick approach.

I thought its a good question however it turned to be a tough one. _________________

Thats correct, even I used brute force. But can we find some better approach for similar question, where constant may be changed i.e. 64 or exponent may be changed. ?

HI.. a quick appch i can think of.. the eq can be written as... x^6-y^2=127...x^6-y^2=(x^3-y)(x^3+y)....which means 127 is product of two int... but if u look at 127..it is prime number... so there is no pair of positive integers which satisfies the condition ...ANS-0

Find the number of pairs of positive integers (x, y) such that x^6 = y^2 + 127.

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

pls share if u used a quick approach to solve this

chetan2u's approach above is excellent, though there's one problem with the analysis. We have

(x^3 + y)(x^3 - y) = 127

so x^3 + y and x^3 - y must be factors of 127. Since 127 is prime, the only possibility is that x^3 + y = 127 and x^3 - y = 1. Now, x cannot be greater than 5, since that would make x^3 larger than 127, and since x^3 must be greater than y, x could only be 4 or 5. Still testing these values, we do find that x = 4 and y = 63 gives a legitimate solution here, so there is one pair of values that works. _________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

HI.. a quick appch i can think of.. the eq can be written as... x^6-y^2=127...x^6-y^2=(x^3-y)(x^3+y)....which means 127 is product of two int... but if u look at 127..it is prime number... so there is no pair of positive integers which satisfies the condition ...ANS-0

This is definitely a tough one... but I do not think the answer is 0.

chetan2u had a great approach, same as the one I used but since 127 is a prime number I went a step further: (x^3-y)(x^3+y)=127 (x^3-y)(x^3+y)=(1)(127)

From here I plugged in different values of x and y to see if I can get (x^3-y) to equal 1 and (x^3+y) to equal 127. Because we know that (x^3-y) must equal 1, y must be x^3-1.

x=1, not possible x=2, not possible x=3, not possible

hi ianstewart... thanks a lot... should not have skipped my mind but it seems in a hurry, just overlooked it... lesson learnt-give a thought before proceeding to next step..

HI.. a quick appch i can think of.. the eq can be written as... x^6-y^2=127...x^6-y^2=(x^3-y)(x^3+y)....which means 127 is product of two int... but if u look at 127..it is prime number... so there is no pair of positive integers which satisfies the condition ...ANS-0

Thats a good point.

x^6-y^2 = 127 (x^3-y)(x^3+y) = 127 x 1

Since x and y both are +ves, (x^3-y) = 1 and (x^3+y) = 127

As I mentioned earlier, x >2. Now the possibilities for x are <5. So x could be 3 or 4 or 5.

I believe, solving these kinds of questions is key to scoring well on the test. When we never know which question is experimental, a tough one as worse as this can always waste time and lower scores. I have had similar questions and I did not have sufficient practise facing them and lost valuable time. The key is to not panic and try to think at least one step beyond or differently for every 3 to 5 seconds when facing a tough problem. _________________

Find the number of pairs of positive integers (x, y) such that x^6 = y^2 + 127.

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

pls share if u used a quick approach to solve this

Here is what I would do:

First thing that comes to my mind is that 127 is prime. Well then, I can't do anything with it right now. Then I see that there are some squares \((x^3)^2 - y^2 = 127\) Let's say\(x^3 = a\)

\(a^2 - y^2 = 127\) \((a+y)(a-y) = 127*1\) We know that a and y are both positive integers. Therefore, their sum, a+y = 127 and their difference, a-y = 1. It is obvious that a and y must be 64 and 63. (or you can solve the two equations simultaneously to get the values for a and y) If a = 64 = x^3, x must be 4. So there is only one pair of values (4, 63).

The question is straight forward because 127 is prime. You get only one pair of values. If instead, we have a composite number with many factors, we need to find the possible values of a and y and then see which values of a work for us. _________________

Re: Find the number of pairs of positive integers (x, y) such [#permalink]
28 Aug 2014, 23:05

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. _________________

Re: Find the number of pairs of positive integers (x, y) such [#permalink]
05 Sep 2014, 07:38

Is it correct to assume that there can only be one set because the integers are positive and x's exponent is higher than y's? Without changing the equation you can almost see that there can only be one solution. But maybe I am thinking too simply...

Re: Find the number of pairs of positive integers (x, y) such [#permalink]
08 Sep 2014, 21:44

1

This post received KUDOS

Expert's post

logophobic wrote:

Is it correct to assume that there can only be one set because the integers are positive and x's exponent is higher than y's? Without changing the equation you can almost see that there can only be one solution. But maybe I am thinking too simply...

No. 127 is a prime number. Try putting in a composite number. Then see whether you get multiple values. Also, some other prime numbers such as 19 may give no solution. _________________

Re: Find the number of pairs of positive integers (x, y) such [#permalink]
10 Sep 2014, 01:01

IanStewart wrote:

xcusemeplz2009 wrote:

Find the number of pairs of positive integers (x, y) such that x^6 = y^2 + 127.

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

pls share if u used a quick approach to solve this

chetan2u's approach above is excellent, though there's one problem with the analysis. We have

(x^3 + y)(x^3 - y) = 127

so x^3 + y and x^3 - y must be factors of 127. Since 127 is prime, the only possibility is that x^3 + y = 127 and x^3 - y = 1. Now, x cannot be greater than 5, since that would make x^3 larger than 127, and since x^3 must be greater than y, x could only be 4 or 5. Still testing these values, we do find that x = 4 and y = 63 gives a legitimate solution here, so there is one pair of values that works.

From above it would follow that if x & y are positive, x^3-y=1 & x^3+y=127. adding the two equations, 2x^3=128, x^3=64, x=4. since solvable with a unique solution, answer B

gmatclubot

Re: Find the number of pairs of positive integers (x, y) such
[#permalink]
10 Sep 2014, 01:01

Can you teach businessmen to be ethical? : he mind is divided into two parts that sometimes conflict, like a small rider sitting on the back of a very...

HBS: Reimagining Capitalism: Business and Big Problems : Growing income inequality, poor or declining educational systems, unequal access to affordable health care and the fear of continuing economic distress...

Over the last week my Facebook wall has been flooded with most positive, almost euphoric emotions: “End of a fantastic school year”, “What a life-changing year it’s been”, “My...