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x^2-y^2<8, x+y>3

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x^2-y^2<8, x+y>3 [#permalink] New post 29 Jun 2018, 17:37
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\(x^{2}-y^{2}< 8\)

\(x+y>3\)

If x and y are integers in the above inequalities and 0 < y < x, what is the greatest possible value of x?



[Reveal] Spoiler: OA
4
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Re: x^2-y^2<8, x+y>3 [#permalink] New post 30 Jun 2018, 09:09
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Re: x^2-y^2<8, x+y>3 [#permalink] New post 30 Jun 2018, 16:13
Carcass wrote:
The answer should be seven.

The OA??


The answer is 4. I don't understand why it gets 4.
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Re: x^2-y^2<8, x+y>3 [#permalink] New post 30 Jun 2018, 16:52
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What is the source of this question. PS: OA means official answer.

I am not sure if this is a GRE question. THIS IS A BAD QUESTION.

condition 1: x+y>3
condition 2: \(x^{2}-y^{2}< 8\)


We start by picking numbers. So the difference between squares has to be less than 8.

condition 1: x+y>3
condition 2: \(x^{2}-y^{2}< 8\)

For example 1,3 now \(3^2-1^2 = 8\) and \(3+1=4\) condition 1 is statisfied condition 2 is not.

Next pair is 3,4. So \(4^2-3^2 = 7\) and and \(3+4=7\). This satisfies but this may not be the gratest value of x. Try the next pair. 5,4

\(5^2-4^2 = 9\) SO not true.

\(6^2-5^2 = 11\)

\(7^2-6^2 = 13\)


Note that the difference between consecutive squares keep on increasing. So 3,4 is the only viable pair.

HERE is why this is a BAD question. \(y>x\).

Say y=1000 x=500.

\(x+y =1500\) holds

\(x^2-y^2=-750000 < 8\) holds

Now here x can have infinitely many values. x=1 and y=1000 holds so does x=999 and y=1000. Also y can have any value.
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Re: x^2-y^2<8, x+y>3 [#permalink] New post 08 Jul 2018, 04:59
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We have x+y>3 AND x^2-y^2<8. Therefore x-y< 8/3 (nearly equals 2.63) since x^2-y^2 = (x+y)(x-y).
x and y are integers, plus 0 < y < x then x-y equal z which is an integer, lower than 2.63. We only have (1,2)
Since the smallest number of x and y that product x-y = 2 is x=3 and y=1. But x^2-y^2=8 which is wrong.
Then only 1 case left which is x-y = 1. We can create a table of x and y from then to see the greatest value of x is 4
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Re: x^2-y^2<8, x+y>3 [#permalink] New post 09 Jul 2018, 14:01
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theAlbatross wrote:
\(x^{2}-y^{2}< 8\)

x+y>3

If x and y are integers in the above inequalities and 0 < y < x, what is the greatest possible value of x? _____


Since \(x\) and \(y\) are positive integers, \(x^2\) and \(y^2\) are PERFECT SQUARES, implying that \(x^2-y^2\)is equal to the difference of two perfect squares.
Make list of perfect squares:
1, 4, 9, 16, 25...
Since \(x^2 - y^2 < 8\), the value of x will be maximized if \(x^2\) and \(y^2\) are the perfect squares in blue, which have a difference of 7.
Thus:
\(x^2 = 16\), implying that the greatest possible value of \(x=4\).

The inequality in red is not necessary to solve the problem.
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Re: x^2-y^2<8, x+y>3 [#permalink] New post 04 Nov 2019, 21:06
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Re: x^2-y^2<8, x+y>3 [#permalink] New post 04 Nov 2019, 23:44
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Re: x^2-y^2<8, x+y>3 [#permalink] New post 04 Nov 2019, 23:49
Carcass wrote:
No.

The tag was wrong because the user used it. All and only the questions from powerprep are located here

https://greprepclub.com/forum/gre-power ... -3118.html

If you see a question with the same tag but NOT from the directory above, then it is NOT a pp question. Absolutely.

Regards


Thanks, Although this question basically i found in Magoosh Very Hard section. Its' from Magoosh.

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Re: x^2-y^2<8, x+y>3 [#permalink] New post 06 Nov 2019, 11:37
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Given : x^2-y^2<8
x+y>3

(x+y)*(x-y)<8
since x+y>3
case 1 : x+y = 4 -> x-y = 1 -> x=2.5 not an integer
case 2 : x+y = 5 -> x-y = 1 -> x=3 and y=2
case 3 : x+y = 6 -> x-y =1 ->x=3.5 not an integer
case 4 : x+y=7 -> x-y=1 -> x=4 and y=3.
As question asks for max value of x, hence 4
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Re: x^2-y^2<8, x+y>3 [#permalink] New post 06 Nov 2019, 11:48
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theAlbatross wrote:
\(x^{2}-y^{2}< 8\)

\(x+y>3\)

If x and y are integers in the above inequalities and 0 < y < x, what is the greatest possible value of x?



[Reveal] Spoiler: OA
7


OA answer should be 4 not 7.
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Re: x^2-y^2<8, x+y>3   [#permalink] 06 Nov 2019, 11:48
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