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# M28-50

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Math Expert
Joined: 02 Sep 2009
Posts: 43335

Kudos [?]: 139557 [0], given: 12794

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16 Sep 2014, 00:44
Expert's post
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Difficulty:

95% (hard)

Question Stats:

44% (00:49) correct 56% (01:26) wrong based on 62 sessions

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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

_________________

Kudos [?]: 139557 [0], given: 12794

Math Expert
Joined: 02 Sep 2009
Posts: 43335

Kudos [?]: 139557 [1], given: 12794

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16 Sep 2014, 00:44
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Expert's post
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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.

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Kudos [?]: 139557 [1], given: 12794

Manager
Joined: 08 Jan 2015
Posts: 86

Kudos [?]: 10 [1], given: 53

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23 Jun 2016, 21:48
1
KUDOS
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.

Kudos [?]: 10 [1], given: 53

Intern
Joined: 13 Jul 2016
Posts: 2

Kudos [?]: 1 [1], given: 2

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28 Oct 2016, 12:06
1
KUDOS
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...

Kudos [?]: 1 [1], given: 2

Intern
Joined: 09 Jul 2016
Posts: 17

Kudos [?]: [0], given: 12

GMAT 1: 730 Q50 V39

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29 May 2017, 09:35
I think this is a poor-quality question and I don't agree with the explanation. Question statement is not clear enough.

Kudos [?]: [0], given: 12

Math Expert
Joined: 02 Sep 2009
Posts: 43335

Kudos [?]: 139557 [0], given: 12794

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29 May 2017, 10:10
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.
_________________

Kudos [?]: 139557 [0], given: 12794

Math Expert
Joined: 02 Sep 2009
Posts: 43335

Kudos [?]: 139557 [0], given: 12794

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29 May 2017, 10:12
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.

Can you be more specific? Question is mathematically correct and of GMAT quality.
_________________

Kudos [?]: 139557 [0], given: 12794

Manager
Joined: 17 May 2016
Posts: 154

Kudos [?]: 34 [0], given: 106

Location: India
GMAT 1: 710 Q48 V40
WE: Marketing (Other)

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14 Jun 2017, 19:24
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.

my 2c!

Kudos [?]: 34 [0], given: 106

Intern
Joined: 22 May 2015
Posts: 14

Kudos [?]: 6 [0], given: 349

Location: India
GMAT 1: 650 Q46 V34
GPA: 3.4

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30 Jul 2017, 02:21
I think this is a high-quality question and I agree with explanation.

Kudos [?]: 6 [0], given: 349

Manager
Joined: 16 Jul 2016
Posts: 133

Kudos [?]: 11 [0], given: 15

Location: India
GPA: 4
WE: Brand Management (Retail)

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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.

Kudos [?]: 11 [0], given: 15

Math Expert
Joined: 02 Sep 2009
Posts: 43335

Kudos [?]: 139557 [0], given: 12794

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23 Aug 2017, 10:32
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;
...
_________________

Kudos [?]: 139557 [0], given: 12794

VP
Joined: 26 Mar 2013
Posts: 1374

Kudos [?]: 323 [0], given: 170

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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 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?

Kudos [?]: 323 [0], given: 170

Math Expert
Joined: 02 Sep 2009
Posts: 43335

Kudos [?]: 139557 [0], given: 12794

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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 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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Kudos [?]: 139557 [0], given: 12794

Re: M28-50   [#permalink] 24 Aug 2017, 09:32
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# M28-50

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