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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]  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? _____
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Re: x^2-y^2<8, x+y>3 [#permalink]  30 Jun 2018, 09:09
Expert's post

The OA??
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Re: x^2-y^2<8, x+y>3 [#permalink]  30 Jun 2018, 16:13
Carcass wrote:

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]  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]  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]  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] 09 Jul 2018, 14:01
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