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# M28-58

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Joined: 02 Sep 2009
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30 Dec 2017, 10:09
sandysilva wrote:
grsm wrote:
dear Bunuel,
I got your solution, it really good ! how ever can you please see where i am going wrong in my approach for choice A.

A. |x−2|<2−y
i intent to open the mod and test the values of x and Y. Therefore 2 cases come up while opening the mod

Case 1 x-2>0 i.e x>2
The equation becomes
x-2 < 2-y
x < 4-y
Now according the above inequality and the condition X>2 there is no value of x or y which satisfies the 2<x<4-y

What am i missing here ?

Case 2 x-2<0 i.e x<2
if we analyze this condition carefully the only value that x can take here is 1 coz, 0<x<2 . However the equations becomes,
-x+2 <2-y
x>y
Now this cannot be correct as y does not have any valid +integer value. 0<y<x<2

This is not a optimal approach, but i am keen to know gap do i have in my concepts ?

thanks

Bunuel wrote:
Official Solution:

(1) $$|x - 2| \lt 2 - y$$. The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus $$0 \lt 2 - y$$, or $$y \lt 2$$ (if $$y$$ is more than or equal to 2, then $$y-2 \leq 0$$ and it cannot be greater than $$|x - 2|$$). Next, since given that $$y$$ is a positive integer, then $$y=1$$.

So, we have that: $$|x - 2| \lt 1$$, which implies that $$-1 \lt x-2 \lt 1$$, or $$1 \lt x \lt 3$$, thus $$x=2=prime$$. Sufficient.

(2) $$x + y - 3 = |1-y|$$. Since $$y$$ is a positive integer, then $$1-y \leq 0$$, thus $$|1-y|=-(1-y)$$. So, we have that $$x + y - 3 = -(1-y)$$, which gives $$x=2=prime$$. Sufficient.

I have the same query.Can any experts please address what GRSM said. I used the same approach.

Thanks

The problem with that solution is that it does not consider the case when x = 2. When you open the modules for |x - 2| you should consider the case x <= 2 and x > 2 OR x < 2 and x >= 2, so you should include = sign for 2 either in the first case or in second. If you do, then you'll find that x = 2 and y = 1 are possible.

Hope it's clear.
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18 Sep 2018, 12:24
Superb Question! Tests concepts very well.
If one is familiar with the concept, it can be solved under 2 minutes.
Re: M28-58 &nbs [#permalink] 18 Sep 2018, 12:24

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# M28-58

Moderators: chetan2u, Bunuel

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