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M28-50

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M28-50 [#permalink]

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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.
[Reveal] Spoiler: OA

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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 \(x\)?

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: M28-50 [#permalink]

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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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I think this is a poor-quality 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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Re: M28-50 [#permalink]

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

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New post 29 May 2017, 11:10
siddhanthsivaraman wrote:
I think this is a poor-quality 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 = 2 and 2^2 = 4 or 5^2 = 25 and 6^2 = 36.
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New post 14 Jun 2017, 20:24
Vikram_Katti wrote:
I think this is a poor-quality 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.

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New post 30 Jul 2017, 03:21
I think this is a high-quality question and I agree with explanation.

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New post 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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New post 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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Re: M28-50 [#permalink]

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New post 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 Bunuel

As 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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New post 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 Bunuel

As 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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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.


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Re: M28-50   [#permalink] 24 Aug 2017, 10:32
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