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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