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The difference between the squares of two positive integers is 2011

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The difference between the squares of two positive integers is 2011  [#permalink]

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New post 11 Apr 2018, 11:17
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The difference between the squares of two positive integers is 2011 what is the greatest possible sum of those two integers?
A) 1002
B) 1005
C) 1007
D) 1809
E) None of these


source: TIME Mock GMAT
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The difference between the squares of two positive integers is 2011  [#permalink]

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New post Updated on: 26 Jun 2018, 06:16
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hetmavani wrote:
The difference between the squares of two positive integers is 2011 what is the greatest possible sum of those two integers?
A) 1002
B) 1005
C) 1007
D) 1809
E) None of these


source: TIME Mock GMAT


Ans : E
Given in the question \(x^2-y^2=2011\)
\((x+y)(x-y)=2011\)
For maximum possible value of \((x+y)\), \((x-y)\)\(\) should be minimum.
x cannot be equal to y , then minimum value of \((x-y)\) is \(1\).
So it gives Maximum value \((x+y)=2011\).

solving \(x-y=1\) and \(x+y=2011\), we get \(x=1006, y =1005\)
so using x-y=1,we get value of x and y as integers, which matches the information given in question stem.
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Originally posted by Princ on 11 Apr 2018, 12:08.
Last edited by Princ on 26 Jun 2018, 06:16, edited 1 time in total.
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The difference between the squares of two positive integers is 2011  [#permalink]

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New post 11 Apr 2018, 12:18
1
hetmavani wrote:
The difference between the squares of two positive integers is 2011 what is the greatest possible sum of those two integers?
A) 1002
B) 1005
C) 1007
D) 1809
E) None of these


source: TIME Mock GMAT


The difference between the square of two positive integers is the product
of the sum and difference of the integers. \(x^2 - y^2 = (x + y)(x - y)\)

2011 = 2011*1(when prime-factorized)
Also, neither of the answer options available divide 2011.

Therefore, Option E(None of these) is the only answer possible!
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Re: The difference between the squares of two positive integers is 2011  [#permalink]

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New post 11 Apr 2018, 14:32
[quote="hetmavani"]The difference between the squares of two positive integers is 2011 what is the greatest possible sum of those two integers?
A) 1002
B) 1005
C) 1007
D) 1809
E) None of these/quote]

(x+y)(x-y)=2011
as 2011 is prime,
x+y=2011
x-y=1
adding,
2x=2012
x=1006
subtracting,
2y=2010
y=1005
1006+1005=2011
E
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Re: The difference between the squares of two positive integers is 2011  [#permalink]

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New post 22 Apr 2018, 20:47
X^2-Y^2=(x+y)(x-y)

X+y max means x-y min
So x-yhas to be equal to 1

So x+y=2011

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Re: The difference between the squares of two positive integers is 2011  [#permalink]

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New post 24 Jun 2018, 16:58
pushpitkc wrote:
hetmavani wrote:
The difference between the squares of two positive integers is 2011 what is the greatest possible sum of those two integers?
A) 1002
B) 1005
C) 1007
D) 1809
E) None of these


source: TIME Mock GMAT


The difference between the square of two positive integers is the product
of the sum and difference of the integers. \(x^2 - y^2 = (x + y)(x - y)\)

2011 = 2011*1(when prime-factorized)
Also, neither of the answer options available divide 2011.

Therefore, Option E(None of these) is the only answer possible!


Hi pushpitkc, is there a quick way to determine that 2011 is prime? I'm only aware of the method where we check every prime less than \(\sqrt{2011}\) but if we do that, the question easily takes 5+ mins.
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Re: The difference between the squares of two positive integers is 2011  [#permalink]

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New post 24 Jun 2018, 22:45
aserghe1 wrote:
pushpitkc wrote:
hetmavani wrote:
The difference between the squares of two positive integers is 2011 what is the greatest possible sum of those two integers?
A) 1002
B) 1005
C) 1007
D) 1809
E) None of these


source: TIME Mock GMAT


The difference between the square of two positive integers is the product
of the sum and difference of the integers. \(x^2 - y^2 = (x + y)(x - y)\)

2011 = 2011*1(when prime-factorized)
Also, neither of the answer options available divide 2011.

Therefore, Option E(None of these) is the only answer possible!


Hi pushpitkc, is there a quick way to determine that 2011 is prime? I'm only aware of the method where we check every prime less than \(\sqrt{2011}\) but if we do that, the question easily takes 5+ mins.


Hi aserghe1

Unfortunately, that is the shortest method to find whether 2011 is a prime number or not.

However, in this case, we needn't necessarily go to the extent of finding whether it is prime.
We just need to check if the prime numbers in the answer options divide 2011. For instance,
take Option A(1002), which can be prime factorized as 2*3*167. If either of the prime factors
do not divide 2001, we can eliminate that particular option.

Hope this helps you!
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Re: The difference between the squares of two positive integers is 2011  [#permalink]

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New post 25 Jun 2018, 15:59
Quote:
Quote:
Hi pushpitkc, is there a quick way to determine that 2011 is prime? I'm only aware of the method where we check every prime less than \(\sqrt{2011}\) but if we do that, the question easily takes 5+ mins.


Hi aserghe1

Unfortunately, that is the shortest method to find whether 2011 is a prime number or not.

However, in this case, we needn't necessarily go to the extent of finding whether it is prime.
We just need to check if the prime numbers in the answer options divide 2011. For instance,
take Option A(1002), which can be prime factorized as 2*3*167. If either of the prime factors
do not divide 2001, we can eliminate that particular option.

Hope this helps you!


I think you're onto something pushpitkc, but I don't understand the logic behind checking if the prime numbers in the answer options divide 2011. Going with your example, if we take Option A (1002), find its prime factorization (2*3*167)... we see that 2011 is not divisible by 2, so eliminate Option A. How does 2011 not being divisible by 2 allow us to eliminate Option A?
Re: The difference between the squares of two positive integers is 2011 &nbs [#permalink] 25 Jun 2018, 15:59
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