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M28-50

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New post 16 Sep 2014, 01:44
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New post 16 Sep 2014, 01:44
1
4
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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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
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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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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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Re: M28-50  [#permalink]

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New post 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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New post 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 Integer

First 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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New post 19 Jun 2018, 02:05
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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New post 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 .

Bunuel


It 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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New post 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 .

Bunuel


It 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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New post 29 Aug 2018, 10:08
I think this is a high-quality 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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New post 29 Aug 2018, 10:18
saplogics wrote:
I think this is a high-quality 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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New post 29 Aug 2018, 13:18
Got it. Thanks for Opening my eyes. Thank you so much. :thumbup:
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New post 16 Dec 2018, 22:45
Bunuel...
Reg Statement (2)..shouldnt we need to consider that Root(X) and Root(y) can also be the negative numbers...thus rendering the Statement insufficient?
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New post 16 Dec 2018, 23:47
Debashis Roy wrote:
Bunuel...
Reg Statement (2)..shouldnt we need to consider that Root(X) and Root(y) can also be the negative numbers...thus rendering the Statement insufficient?


First of all, (2) says that \(\sqrt{x}\) and \(\sqrt{y}\) are prime numbers. Only positive numbers can be primes.

Next, when the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).
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New post 27 Dec 2019, 21:53
I think this is a poor-quality question and I don't agree with the explanation. x=16,y=25 satisfies the stem and (1)

why do we need to restrict x and y to 2,3?
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New post 28 Dec 2019, 01:14
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