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# 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  [#permalink]

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12 Nov 2019, 01:33
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35% (medium)

Question Stats:

72% (01:59) correct 28% (02:20) wrong based on 129 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

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The difference between the squares of two positive integers is 2011 wh  [#permalink]

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12 Nov 2019, 04:21
Bunuel 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

Are You Up For the Challenge: 700 Level Questions

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
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Re: The difference between the squares of two positive integers is 2011 wh  [#permalink]

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20 Nov 2019, 18:14
Bunuel 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

Are You Up For the Challenge: 700 Level Questions

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.

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Re: The difference between the squares of two positive integers is 2011 wh  [#permalink]

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24 Nov 2019, 20:05

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.

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The difference between the squares of two positive integers is 2011 wh  [#permalink]

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17 Jan 2020, 03:16
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
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Re: The difference between the squares of two positive integers is 2011 wh  [#permalink]

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17 Jan 2020, 10:36
Bunuel 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

Are You Up For the Challenge: 700 Level Questions

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

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)
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Re: The difference between the squares of two positive integers is 2011 wh  [#permalink]

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17 Jan 2020, 10:56
[quote="Bunuel"]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?
Re: The difference between the squares of two positive integers is 2011 wh   [#permalink] 17 Jan 2020, 10:56