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The difference between the squares of two positive integers is 2011

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Intern
Joined: 24 Mar 2018
Posts: 8
The difference between the squares of two positive integers is 2011 [#permalink]

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11 Apr 2018, 11:17
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Difficulty:

55% (hard)

Question Stats:

63% (01:09) correct 38% (01:20) wrong based on 16 sessions

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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
Manager
Joined: 22 Feb 2018
Posts: 77
The difference between the squares of two positive integers is 2011 [#permalink]

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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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Joined: 26 Feb 2016
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The difference between the squares of two positive integers is 2011 [#permalink]

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11 Apr 2018, 12:18
1
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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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Director
Joined: 07 Dec 2014
Posts: 963
Re: The difference between the squares of two positive integers is 2011 [#permalink]

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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
2x=2012
x=1006
subtracting,
2y=2010
y=1005
1006+1005=2011
E
Manager
Joined: 02 Oct 2017
Posts: 55
Re: The difference between the squares of two positive integers is 2011 [#permalink]

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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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