Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 09 May 2009
Posts: 151

Find the number of pairs of positive integers (x, y) such
[#permalink]
Show Tags
Updated on: 03 Jul 2013, 01:13
Question Stats:
36% (02:20) correct 64% (02:26) wrong based on 298 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by xcusemeplz2009 on 11 Dec 2009, 22:23.
Last edited by Bunuel on 03 Jul 2013, 01:13, edited 1 time in total.
Renamed the topic, edited the question and added the OA.




Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9869
Location: Pune, India

Re: x^6 = y^2 + 127
[#permalink]
Show Tags
21 Oct 2011, 03:34
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 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\(x^3 = a\) \(a^2  y^2 = 127\) \((a+y)(ay) = 127*1\) We know that a and y are both positive integers. Therefore, their sum, a+y = 127 and their difference, ay = 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.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Math Expert
Joined: 02 Aug 2009
Posts: 8302

Find the number of pairs of positive integers (x, y) such
[#permalink]
Show Tags
12 Dec 2009, 07:16
HI.. a quick appch i can think of.. the eq can be written as... \(x^6y^2=127...x^6y^2=(x^3y)(x^3+y)=127.\)...which means 127 is product of two integer. but if u look at 127..it is prime number... 127=1*127=\((x^3y)(x^3+y)\) x^3y=1 and x^3+y=127...Add both to get 2*x^3=128, so x^3=64, therefore, X=4 and y =641=63.. so, there is only one pair of positive integers which satisfies the condition 1*127 ... ANS1
_________________




SVP
Joined: 29 Aug 2007
Posts: 1795

Re: x^6 = y^2 + 127
[#permalink]
Show Tags
12 Dec 2009, 00:06
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 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(1127). y is an irrational number. Not possible. If x = 2, x^6 = 64 and y = sqrt(64127). y is an irrational number. Not possible. If x = 3, x^6 = 729 and y = sqrt(729127) = 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 25 minuets. Until and unless there is a quick approach, I would say this not a gmattype 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.



Intern
Joined: 12 Oct 2008
Posts: 35

Re: x^6 = y^2 + 127
[#permalink]
Show Tags
12 Dec 2009, 01:48
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. ?



GMAT Tutor
Joined: 24 Jun 2008
Posts: 1829

Re: x^6 = y^2 + 127
[#permalink]
Show Tags
12 Dec 2009, 08:03
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.
_________________
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



Manager
Joined: 13 Aug 2009
Posts: 120

Re: x^6 = y^2 + 127
[#permalink]
Show Tags
12 Dec 2009, 08:06
chetan2u wrote: HI.. a quick appch i can think of.. the eq can be written as... x^6y^2=127...x^6y^2=(x^3y)(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 ...ANS0 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^3y)(x^3+y)=127 (x^3y)(x^3+y)=(1)(127) From here I plugged in different values of x and y to see if I can get (x^3y) to equal 1 and (x^3+y) to equal 127. Because we know that (x^3y) must equal 1, y must be x^31. x=1, not possible x=2, not possible x=3, not possible x=4, y=63 (x^3y)(x^3+y)=127 (4^363)*(4^3+63)=127 x=5, not possible So answer is B. 1 Can someone please verify the OA?



Math Expert
Joined: 02 Aug 2009
Posts: 8302

Re: x^6 = y^2 + 127
[#permalink]
Show Tags
12 Dec 2009, 08:25
hi ianstewart... thanks a lot... should not have skipped my mind but it seems in a hurry, just overlooked it... lesson learntgive a thought before proceeding to next step..
_________________



SVP
Joined: 29 Aug 2007
Posts: 1795

Re: x^6 = y^2 + 127
[#permalink]
Show Tags
12 Dec 2009, 13:10
chetan2u wrote: HI.. a quick appch i can think of.. the eq can be written as... x^6y^2=127...x^6y^2=(x^3y)(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 ...ANS0 Thats a good point. x^6y^2 = 127 (x^3y)(x^3+y) = 127 x 1 Since x and y both are +ves, (x^3y) = 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. After few trails, x = 4. Agree with OA as B.



Senior Manager
Joined: 21 Jul 2009
Posts: 282
Schools: LBS, INSEAD, IMD, ISB  Anything with just 1 yr program.

Re: x^6 = y^2 + 127
[#permalink]
Show Tags
12 Dec 2009, 15:21
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.



Intern
Joined: 03 Sep 2014
Posts: 6

Re: Find the number of pairs of positive integers (x, y) such
[#permalink]
Show Tags
05 Sep 2014, 08:38
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...



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9869
Location: Pune, India

Re: Find the number of pairs of positive integers (x, y) such
[#permalink]
Show Tags
08 Sep 2014, 22:44
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.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 25 Aug 2014
Posts: 3
Location: India
Concentration: Finance, International Business
GPA: 2.95
WE: Military Officer (Military & Defense)

Re: Find the number of pairs of positive integers (x, y) such
[#permalink]
Show Tags
10 Sep 2014, 02:01
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^3y=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



Intern
Joined: 12 Nov 2015
Posts: 43

Re: Find the number of pairs of positive integers (x, y) such
[#permalink]
Show Tags
21 Dec 2015, 23:44
It could have been a bit simple , if they had mentioned that x and y are consecutive integers.



Intern
Joined: 28 Nov 2015
Posts: 1

Re: Find the number of pairs of positive integers (x, y) such
[#permalink]
Show Tags
22 Dec 2015, 04:03
VeritasPrepKarishma 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 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\(x^3 = a\) \(a^2  y^2 = 127\) \((a+y)(ay) = 127*1\) We know that a and y are both positive integers. Therefore, their sum, a+y = 127 and their difference, ay = 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. i am unable to understand , how did we come to the part (a+y)= 127 and (ay)=1



Math Expert
Joined: 02 Aug 2009
Posts: 8302

Re: Find the number of pairs of positive integers (x, y) such
[#permalink]
Show Tags
22 Dec 2015, 04:14
onewaysuccess wrote: VeritasPrepKarishma 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 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\(x^3 = a\) \(a^2  y^2 = 127\) \((a+y)(ay) = 127*1\) We know that a and y are both positive integers. Therefore, their sum, a+y = 127 and their difference, ay = 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. i am unable to understand , how did we come to the part (a+y)= 127 and (ay)=1 Hi, we know x and y are +ive integers and x^3=a.. therefore x^6 = y^2 + 127 can be written as a^2=y^2+127... a^2y^2=127.. a^2y^2=(ay)(a+y)=127 127 can be written as 127*1 so (ay)(a+y)=127*1.. so a+y=127 and ay=1.. ay=1 shows a and y are consecutive numbers and we can find from the two eq as 63 and 64.. x^3=a=64.. so x=4.. ans 4 and 63 hope it is clear
_________________



Math Expert
Joined: 02 Sep 2009
Posts: 59674

Re: Find the number of pairs of positive integers (x, y) such
[#permalink]
Show Tags
22 Dec 2015, 04:15
onewaysuccess wrote: VeritasPrepKarishma 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 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\(x^3 = a\) \(a^2  y^2 = 127\) \((a+y)(ay) = 127*1\) We know that a and y are both positive integers. Therefore, their sum, a+y = 127 and their difference, ay = 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. i am unable to understand , how did we come to the part (a+y)= 127 and (ay)=1 127 is a prime number, and it can be broken into the product of two positive integers only in one way: 127 = 1*127.
_________________



Intern
Status: GMAT in August 2018
Joined: 05 Mar 2018
Posts: 45
Location: India
Concentration: Leadership, Strategy
WE: Law (Consulting)

Re: Find the number of pairs of positive integers (x, y) such
[#permalink]
Show Tags
18 Apr 2018, 23:26
Dear Experts, Can we solve this as below:
We are given that x^6−y^2=127... or (x^3)^2 − y^2 = 127 [in the normal A^2B^2 format]. Without knowing that 127 is a prime or not, we know that x & y can assume 4 values  i.e. x or +x and y or +y, yet the equation will remain same as both as squares and will always be +ive. Thus we get 4 different pairs of x & y. But the questions requires only +ive pairs, which is only one!
So ans is 1 pair.. Works? Any flaws? Thanks.




Re: Find the number of pairs of positive integers (x, y) such
[#permalink]
18 Apr 2018, 23:26






