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The difference between the squares of two positive integers is 2011

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Joined: 24 Mar 2018
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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
[Reveal] Spoiler: OA
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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, 12:08
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 \((x+y)=2011\).

Additional: solving \(x-y=1\) and \(x+y=2011\), we get \(x=1006, y =1005\)
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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
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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


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] 22 Apr 2018, 20:47
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The difference between the squares of two positive integers is 2011

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