The difference between the squares of two positive integers is 2011 wh

Question

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

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chetan2u · Answer

Let the two numbers be a and b, so  a^2-b^2=2011....(a-b)(a+b)=1*2011 
 a-b=1  and  a+b=2011 ...
Straight a+b=2011 
OR
Add to get 2a=1+2011...a=1006 and b, therefore,=1005
 SUM=1005+1006

E

ScottTargetTestPrep · Answer

We can create the equation:

a^2 - b^2 = 2011

(a + b)(a - b) = 2011

Since 2011 is a prime, we see that a + b = 2011 and a - b = 1. We see that the sum of the two integers must be 2011.

Answer: E

EgmatQuantExpert · Answer

Solution  
   Given:    
 • The difference between the squares of two positive integers = 2011

To find:    
 • The greatest possible sum of those two integers

Approach and Working Out:    
 • Given,  a^2 – b^2 = 2011 
 o Implies, (a + b) * (a – b) = 2011

• We know that 2011 is a prime number, so it can be expressed as product of two positive integers in only one way = 1 * 2011
• Thus, a + b = 2011

Hence, the correct answer is Option E.

Answer: E

LevanKhukhunashvili · Answer

We are given
 a^2-b^2=2011  -->  (a-b)*(a+b)=2011 
We want to maximise a+b

I don't know whether 2011 is prime or not
The only thing I know is that it's nor divisible by 2, 3, 5

What to do
Let us glance at the options

1) 1002; we cant multiply 1002 by any integer value to get 2011 (we are looking for integer value because the stem says a and b are integers, so their subtraction would be an integer as well)
B. 1005; nope
C. 1007; no way
D. 1809; definitely no
E. None of these

We are left with  E

Abhishek009 · Answer

a^2 - b^2 = 2011 
 (a + b)(a - b) = 2011

Divide 2011/2 = 1005.50

Now, Let a = 1006 & b = 1005

Now, we have -

(1006 + 1005)( 1006 - 1005 ) = 2011*1

Thus, the greatest possible sum of those two integers must be 2011 , Answer must be (E)

adamhostetter · Answer

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

Can you solve by process of elim here?

(a^2 - b^2 ) = 2011
(a + b ) (a - b) = 2011
therefore:
(a - b) = 2011 / (a + b) 
and (a - b) must be a integer.

check divisibility of answer choices. They all fail <-- therefore: none of these (E).

Is this acceptable?

rito54 · Answer

Hello @ ScottTargetTestPrep  ! but how do you know that 2011 is a prime number? Thanks Like is this something you learn by heart?

pierjoejoe · Answer

i don't understand why being a prime should be important

x and y are positive integers
if x2 - y2 = 2011
thus, we know that 
(x-y)(x+y) = 2011

to maximize the sum, just minimize the difference. the least integer value for the difference is 1. Thus the sum would be 2011. 
for this reason, the max value for the sum is 2011

why it should matter that 2011 is a prime.
for example, if 
x2 - y2 = 16
than 
(x-y)(x+y) = 16
minimum difference is 1, max sum is 16.

ScottTargetTestPrep 
 Abhishek009 
 Bunuel

satish_sahoo · Answer

x^2-y^2= 2011
(x+y)(x-y)= 2011

To make (x+y) greatest, (x-y) should be 1...or else if we increase the value of (x-y), we will have to decrease the value of (x+y) accordingly.

So, Opt E

The difference between the squares of two positive integers is 2011 wh

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The difference between the squares of two positive integers is 2011 wh

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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My Rewards