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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.
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Re: M2850
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23 Jun 2016, 22:48
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.



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16 Sep 2014, 01:44
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
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Re: M2850
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28 Oct 2016, 13:06
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...



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29 May 2017, 10:35
I think this is a poorquality question and I don't agree with the explanation. Question statement is not clear enough.



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siddhanthsivaraman wrote: 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... Consecutive perfect square are say 1^1 = 1 and 2^2 = 4 or 5^2 = 25 and 6^2 = 36.
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29 May 2017, 11:12
Vikram_Katti wrote: I think this is a poorquality question and I don't agree with the explanation. Question statement is not clear enough. Can you be more specific? Question is mathematically correct and of GMAT quality.
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14 Jun 2017, 20:24
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!



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30 Jul 2017, 03:21
I think this is a highquality question and I agree with explanation.



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23 Aug 2017, 07:49
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.



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23 Aug 2017, 11:32
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; ...
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24 Aug 2017, 08:40
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?



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24 Aug 2017, 10:32
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 \(y=(prime_2)^2=9\).
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Re: M2850
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25 Jan 2018, 05:33
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.
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25 Jan 2018, 05:46
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.
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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 .
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19 Jun 2018, 02:13
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.
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19 Jun 2018, 02:26
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.
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29 Aug 2018, 10:08
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.



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29 Aug 2018, 10:18
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.
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