December 14, 2018 December 14, 2018 10:00 PM PST 11:00 PM PST Carolyn and Brett  nicely explained what is the typical day of a UCLA student. I am posting below recording of the webinar for those who could't attend this session. December 15, 2018 December 15, 2018 07:00 AM PST 09:00 AM PST Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 51215

Question Stats:
43% (01:07) correct 57% (01:40) wrong based on 113 sessions
HideShow timer Statistics



Math Expert
Joined: 02 Sep 2009
Posts: 51215

Re M2850
[#permalink]
Show Tags
16 Sep 2014, 00:44
Official Solution:If \(0 \lt x \lt y\) and \(x\) and \(y\) are consecutive perfect squares, what is the remainder when \(y\) is divided by \(x\)? Notice that since \(x\) and \(y\) are consecutive perfect squares, then \(\sqrt{x}\) and \(\sqrt{y}\) are consecutive integers. (1) Both \(x\) and \(y\) have 3 positive factors. This statement implies that \(x=(prime_1)^2\) and \(y=(prime_2)^2\). From above we have that \(\sqrt{x}=prime_1\) and \(\sqrt{y}=prime_2\) are consecutive integers. The only two consecutive integers which are primes are 2 and 3. Thus, \(x=(prime_1)^2=4\) and \(y=(prime_2)^2=9\). The remainder when 9 is divided by 4 is 1. Sufficient. (2) Both \(\sqrt{x}\) and \(\sqrt{y}\) are prime numbers. The same here: \(\sqrt{x}=2\) and \(\sqrt{y}=3\). Sufficient. Answer: D
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Current Student
Joined: 08 Jan 2015
Posts: 79

Re: M2850
[#permalink]
Show Tags
23 Jun 2016, 21:48
I think it's a low quality question, since there is no commonly used definition for consecutive perfect squares. The correct definition should be  perfect squares of consecutive integers. Otherwise the only pair of consecutive perfect squares is 0^2 and 1^2. https://proofwiki.org/wiki/Zero_and_One_are_the_only_Consecutive_Perfect_Squares. Besides that, If the question implies that 36 are 49 are consecutive perfect squares (since it's 6^2 and 7^2), then I don't see a reason, why should one not consider 7 and 11 as consecutive primes. Which then makes the correct answer choice E, because 1 and 2 are tautological.



Intern
Joined: 13 Jul 2016
Posts: 2

Re: M2850
[#permalink]
Show Tags
28 Oct 2016, 12:06
I think this is a poorquality question and I don't agree with the explanation. This is a poor quality question and meaning of consecutive perfect squares is not clear while attempting...



Intern
Joined: 09 Jul 2016
Posts: 17

Re: M2850
[#permalink]
Show Tags
29 May 2017, 09:35
I think this is a poorquality question and I don't agree with the explanation. Question statement is not clear enough.



Math Expert
Joined: 02 Sep 2009
Posts: 51215

Re: M2850
[#permalink]
Show Tags
29 May 2017, 10:10



Math Expert
Joined: 02 Sep 2009
Posts: 51215

Re: M2850
[#permalink]
Show Tags
29 May 2017, 10:12



Manager
Joined: 17 May 2016
Posts: 182
Location: India
WE: Marketing (Education)

Re: M2850
[#permalink]
Show Tags
14 Jun 2017, 19:24
Vikram_Katti wrote: I think this is a poorquality question and I don't agree with the explanation. Question statement is not clear enough. I think in a hurry to solve the problem, it was assumed that the numbers are consecutives primes, and not just consecutive numbers. When actually we need to find consecutives numbers that are prime. Hence the mistake, which I also committed and was baffled at the explanation. On second thoughts, its a clear cut question, absolutely GMAT style. my 2c!



Intern
Joined: 22 May 2015
Posts: 14
Location: India
GPA: 3.4

Re: M2850
[#permalink]
Show Tags
30 Jul 2017, 02:21
I think this is a highquality question and I agree with explanation.



Manager
Joined: 16 Jul 2016
Posts: 246
Location: India
GPA: 4
WE: Brand Management (Retail)

Re: M2850
[#permalink]
Show Tags
23 Aug 2017, 06:49
poor quality question
From above we have that x√=prime1x=prime1 and y√=prime2y=prime2 are consecutive integers. How was this proved ? Expect a better explanation.



Math Expert
Joined: 02 Sep 2009
Posts: 51215

Re: M2850
[#permalink]
Show Tags
23 Aug 2017, 10:32



SVP
Joined: 26 Mar 2013
Posts: 1911

Re: M2850
[#permalink]
Show Tags
24 Aug 2017, 07:40
Bunuel wrote: Yashkumar wrote: poor quality question
From above we have that x√=prime1x=prime1 and y√=prime2y=prime2 are consecutive integers. How was this proved ? Expect a better explanation. x and y are consecutive perfect squares, so x and y could be: \(x = 1\) and \(y = 4\) > \(\sqrt{x}=1\) and \(\sqrt{y}=2\), consecutive integers; \(x = 4\) and \(y = 9\) > \(\sqrt{x}=2\) and \(\sqrt{y}=3\), consecutive integers; \(x = 9\) and \(y = 16\) > \(\sqrt{x}=3\) and \(\sqrt{y}=4\), consecutive integers; ... Dear BunuelAs u stated above the consecutive perfect square 1, 4, 9, 16, 25. From statement 1, it is invalid to consider 9 & 16 because both are consecutive but 16 does not consist of 3 factors. Am I right?



Math Expert
Joined: 02 Sep 2009
Posts: 51215

Re: M2850
[#permalink]
Show Tags
24 Aug 2017, 09:32
Mo2men wrote: Bunuel wrote: Yashkumar wrote: poor quality question
From above we have that x√=prime1x=prime1 and y√=prime2y=prime2 are consecutive integers. How was this proved ? Expect a better explanation. x and y are consecutive perfect squares, so x and y could be: \(x = 1\) and \(y = 4\) > \(\sqrt{x}=1\) and \(\sqrt{y}=2\), consecutive integers; \(x = 4\) and \(y = 9\) > \(\sqrt{x}=2\) and \(\sqrt{y}=3\), consecutive integers; \(x = 9\) and \(y = 16\) > \(\sqrt{x}=3\) and \(\sqrt{y}=4\), consecutive integers; ... Dear BunuelAs u stated above the consecutive perfect square 1, 4, 9, 16, 25. From statement 1, it is invalid to consider 9 & 16 because both are consecutive but 16 does not consist of 3 factors. Am I right? We can consider the above values from the stem. From (1), the only values possible are \(x=(prime_1)^2=4\) and \(y=(prime_2)^2=9\).
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



General GMAT Forum Moderator
Joined: 29 Jan 2015
Posts: 1046
Location: India
WE: General Management (NonProfit and Government)

Re: M2850
[#permalink]
Show Tags
25 Jan 2018, 04:33
Bunuel wrote: Official Solution:
If \(0 \lt x \lt y\) and \(x\) and \(y\) are consecutive perfect squares, what is the remainder when \(y\) is divided by \(x\)? Notice that since \(x\) and \(y\) are consecutive perfect squares, then \(\sqrt{x}\) and \(\sqrt{y}\) are consecutive integers. (1) Both \(x\) and \(y\) have 3 positive factors. This statement implies that \(x=(prime_1)^2\) and \(y=(prime_2)^2\). From above we have that \(\sqrt{x}=prime_1\) and \(\sqrt{y}=prime_2\) are consecutive integers. The only two consecutive integers which are primes are 2 and 3. Thus, \(x=(prime_1)^2=4\) and \(y=(prime_2)^2=9\). The remainder when 9 is divided by 4 is 1. Sufficient. (2) Both \(\sqrt{x}\) and \(\sqrt{y}\) are prime numbers. The same here: \(\sqrt{x}=2\) and \(\sqrt{y}=3\). Sufficient.
Answer: D Hi Bunuel, In the above question Statement 1 says that both x and y have 3 positive factor and you have written that the statement implies that x=(prime1)2x=(prime1)2 and y=(prime2)2y=(prime2)2. How did you arrive at this? Can you please elaborate? Thanks.
_________________
If you liked my post, kindly give me a Kudos. Thanks.



Math Expert
Joined: 02 Sep 2009
Posts: 51215

Re: M2850
[#permalink]
Show Tags
25 Jan 2018, 04:46
rohan2345 wrote: Bunuel wrote: Official Solution:
If \(0 \lt x \lt y\) and \(x\) and \(y\) are consecutive perfect squares, what is the remainder when \(y\) is divided by \(x\)? Notice that since \(x\) and \(y\) are consecutive perfect squares, then \(\sqrt{x}\) and \(\sqrt{y}\) are consecutive integers. (1) Both \(x\) and \(y\) have 3 positive factors. This statement implies that \(x=(prime_1)^2\) and \(y=(prime_2)^2\). From above we have that \(\sqrt{x}=prime_1\) and \(\sqrt{y}=prime_2\) are consecutive integers. The only two consecutive integers which are primes are 2 and 3. Thus, \(x=(prime_1)^2=4\) and \(y=(prime_2)^2=9\). The remainder when 9 is divided by 4 is 1. Sufficient. (2) Both \(\sqrt{x}\) and \(\sqrt{y}\) are prime numbers. The same here: \(\sqrt{x}=2\) and \(\sqrt{y}=3\). Sufficient.
Answer: D Hi Bunuel, In the above question Statement 1 says that both x and y have 3 positive factor and you have written that the statement implies that x=(prime1)2x=(prime1)2 and y=(prime2)2y=(prime2)2. How did you arrive at this? Can you please elaborate? Thanks. Finding the Number of Factors of an IntegerFirst make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers. The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself. Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\) Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors. So, according to the above, an integer to have 3 factor it must be \(x=(prime)^2\). In this case the number of factors = (2 + 1) = 3: 1, prime and prime^2. For example, 2^2 has three factors 1, 2, and 4; 3^2 has three factors 1, 3, and 9.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 26 Feb 2018
Posts: 58
WE: Sales (Internet and New Media)

Bunuel wrote: If \(0 \lt x \lt y\) and \(x\) and \(y\) are consecutive perfect squares, what is the remainder when \(y\) is divided by \(x\)?
(1) Both \(x\) and \(y\) have 3 positive factors.
(2) Both \(\sqrt{x}\) and \(\sqrt{y}\) are prime numbers. D was very prominent , but I overthought , 1st statement : 3 positive factors , this means the number is prime . But in this case , 5 & 11 can also be a option ? even 5 & 7 ? If y is 7 , and x is 5 then the remainder is 2 , and if y is 3 and x is 2 , then the remainder is 1 . This is the only confusion I'm having . other that both the statement clearly indicates that x and y are prime . Statement 2 , clearly says , that x & y are prime , as both are perfect squares I chose E , as I overlooked that X & Y are consecutive perfect square. I was determined that X & Y both are prime, Now when I'm reading it again , it bloody says consecutive perfect square , hence 2 & 3 are the only option. Makes perfect sense. Nice question Bunuel .
_________________
" Can't stop learning and failing"



Math Expert
Joined: 02 Sep 2009
Posts: 51215

Re: M2850
[#permalink]
Show Tags
19 Jun 2018, 01:13
loserunderachiever wrote: Bunuel wrote: If \(0 \lt x \lt y\) and \(x\) and \(y\) are consecutive perfect squares, what is the remainder when \(y\) is divided by \(x\)?
(1) Both \(x\) and \(y\) have 3 positive factors.
(2) Both \(\sqrt{x}\) and \(\sqrt{y}\) are prime numbers. D was very prominent , but I overthought , 1st statement : 3 positive factors , this means the number is prime . But in this case , 5 & 11 can also be a option ? even 5 & 7 ? If y is 7 , and x is 5 then the remainder is 2 , and if y is 3 and x is 2 , then the remainder is 1 . This is the only confusion I'm having . other that both the statement clearly indicates that x and y are prime . Can you please explain the remainder keeps on changing if we are considering other primes except 2 and 3 . I chose E , I was determined that I should be D , but as the remainder was changing I chose E . BunuelIt seems that you are missing the crucial part the stem tells us: \(x\) and \(y\) are consecutive perfect squares. So, for example: 1^2 and 2^2; 2^2 and 3^2; 3^2 and 4^2; ... So, if \(x\) and \(y\) are consecutive perfect squares, then \(\sqrt{x}\) and \(\sqrt{y}\) are consecutive integers: 1 and 2; 2 and 3; 3 and 4; ... Both statements imply that \(\sqrt{x}\) and \(\sqrt{y}\) are primes. The only two consecutive integers which are primes are 2 and 3.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 26 Feb 2018
Posts: 58
WE: Sales (Internet and New Media)

Re: M2850
[#permalink]
Show Tags
19 Jun 2018, 01:26
Bunuel wrote: loserunderachiever wrote: Bunuel wrote: If \(0 \lt x \lt y\) and \(x\) and \(y\) are consecutive perfect squares, what is the remainder when \(y\) is divided by \(x\)?
(1) Both \(x\) and \(y\) have 3 positive factors.
(2) Both \(\sqrt{x}\) and \(\sqrt{y}\) are prime numbers. D was very prominent , but I overthought , 1st statement : 3 positive factors , this means the number is prime . But in this case , 5 & 11 can also be a option ? even 5 & 7 ? If y is 7 , and x is 5 then the remainder is 2 , and if y is 3 and x is 2 , then the remainder is 1 . This is the only confusion I'm having . other that both the statement clearly indicates that x and y are prime . Can you please explain the remainder keeps on changing if we are considering other primes except 2 and 3 . I chose E , I was determined that I should be D , but as the remainder was changing I chose E . BunuelIt seems that you are missing the crucial part the stem tells us: \(x\) and \(y\) are consecutive perfect squares. So, for example: 1^2 and 2^2; 2^2 and 3^2; 3^2 and 4^2; ... So, if \(x\) and \(y\) are consecutive perfect squares, then \(\sqrt{x}\) and \(\sqrt{y}\) are consecutive integers: 1 and 2; 2 and 3; 3 and 4; ... Both statements imply that \(\sqrt{x}\) and \(\sqrt{y}\) are primes. The only two consecutive integers which are primes are 2 and 3. Makes sense ! bloodyhell , I missed the question stem as consecutive perfect squares. Thank you Bunuel . Nice question.
_________________
" Can't stop learning and failing"



Intern
Joined: 07 May 2018
Posts: 4

Re M2850
[#permalink]
Show Tags
29 Aug 2018, 09:08
I think this is a highquality question and I don't agree with the explanation. Hi Bunuel, I went through the explanation and the other comments. I may be wrong but below is the reason I think this needs more discussion. All squares of prime numbers have exactly 3 factors. However the question stem states that the Squares of primes are consecutive. Not that the Prime numbers themselves are consecutive. So we can have any primes picked. 2,3 or 5,7 or anything like 11,13. In all these cases rootX= the prime1 and rootY is the Prime2. But if you see all the different X and Y for these pair of primes will yield different Remainders. This is precisely the reason I think Statement 1 is insufficient. Try 169(y)/121(x) = 48(r). 49(y)/25(x)=24(r).. In that same logic Statement 1 & 2 combined doesn't help getting a fixed answer. Thus I went for choice E. Kindly help me understand If I read something wrong. Thanks for your attention in advance.



Math Expert
Joined: 02 Sep 2009
Posts: 51215

Re: M2850
[#permalink]
Show Tags
29 Aug 2018, 09:18
saplogics wrote: I think this is a highquality question and I don't agree with the explanation. Hi Bunuel, I went through the explanation and the other comments. I may be wrong but below is the reason I think this needs more discussion. All squares of prime numbers have exactly 3 factors. However the question stem states that the Squares of primes are consecutive. Not that the Prime numbers themselves are consecutive. So we can have any primes picked. 2,3 or 5,7 or anything like 11,13. In all these cases rootX= the prime1 and rootY is the Prime2. But if you see all the different X and Y for these pair of primes will yield different Remainders. This is precisely the reason I think Statement 1 is insufficient. Try 169(y)/121(x) = 48(r). 49(y)/25(x)=24(r).. In that same logic Statement 1 & 2 combined doesn't help getting a fixed answer. Thus I went for choice E. Kindly help me understand If I read something wrong. Thanks for your attention in advance. 169 =13^2 and 121 = 11^2 are not consecutive perfect square. 169 = 13^2 and 144 = 12^2 are. Or 144 = 12^2 and 11^2. 49 =7^2 and 121 = 5^2 are not consecutive perfect square.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics







Go to page
1 2
Next
[ 21 posts ]



