If x and y are positive integers and 2013 + x^2 = y^2, what is the

Question

If x and y are positive integers and 2013 + x^2 = y^2, what is the minimum possible value of xy?

A. 8342
B. 671
C. 668
D. 658
E. 645

Oppenheimer1945 · Answer

(y-x)(y+x)=2013

factors(2013)=1.3..11.33.61.183.671.2013

for  min(xy) , choose mid numbers  33,61, 
 x=\frac{(33+61)}{2}, y=\frac{(61-33)}{2} 
 min(xy)=14*47=658

gmatophobia · Answer

The minimum possible value of  xy  occurs  x   and  y  have as close values as possible

x^2 - y^2 = -2013

Prime factorization of  2013 = 3 * 11 * 61

x^2 - y^2 = -3 * 11 * 61

(x+y)(x-y) = -33 * 61

x - y = -33

x + y= 61

Solve for x and y

x = 14

y = 47

x * y = 14 * 47 = 658

Option D

BrushMyQuant · Answer

Given that x and y are positive integers and 2013 + x^2 = y^2. And we need to find the minimum possible value of xy

2013 + x^2 = y^2
 => y^2 - x^2 = 2013 = 33*61
=> (y-x)*(y+x) = 33*61

Now, xy will be minimum when x-y and x+y are close to each other as 
 
  When we subtract them and take an average then we get a very small number and that makes one number smaller and the product small too.  
  When we add them and take an average then we get a a number which is close to these two numbers, which is the other number in the product.

Lets take some examples to understand this  
 
    -    Case 1:    y-x and y+x are far from each other
  
y-x = 1 and y+x = 2013
=> x = 1006 and y = 100
=> xy = 10,13,042 (  very big number  ) 
      Case 2:    y-x and y+x are close to each other
  
=> y-x = 33 and y+x = 61
=> x = 14 and y = 47
=> xy = 658 (  minimum value  )

So,  Answer will be D 
Hope it helps!

bumpbot · Answer

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If x and y are positive integers and 2013 + x^2 = y^2, what is the

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Lets take some examples to understand this
-Case 1: y-x and y+x are far from each other y-x = 1 and y+x = 2013 => x = 1006 and y = 100 => xy = 10,13,042 (very big number)	Case 2: y-x and y+x are close to each other => y-x = 33 and y+x = 61 => x = 14 and y = 47 => xy = 658 (minimum value)

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If x and y are positive integers and 2013 + x^2 = y^2, what is the

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