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M02-25

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New post 15 Sep 2014, 23:18
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New post 15 Sep 2014, 23:18
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Official Solution:


If \(x\) and \(y\) are positive integers, is \(x\) a prime number?

(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 \(0 \lt 2 - y\), thus \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(2-y \le{0}\) 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 \(1-y \le {0}\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.


Answer: D
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Re: M02-25  [#permalink]

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New post 27 Nov 2014, 11:44
[quote="Bunuel"]Official Solution:


(if \(y\) is more than or equal to 2, then \(y-2 \le{0}\)


Since \(y\) is a positive integer, then \(1-y \le {0}\),

I have not understood the above two things. Could you please help me on this? How can we get these equation?
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New post 28 Nov 2014, 04:33
1
Raihanuddin wrote:
Bunuel wrote:
Official Solution:


(if \(y\) is more than or equal to 2, then \(y-2 \le{0}\)


Since \(y\) is a positive integer, then \(1-y \le {0}\),

I have not understood the above two things. Could you please help me on this? How can we get these equation?


I tried to explain this in my solution.

1. (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 \(0 \lt 2 - y\), thus \(y \lt 2\).

2. For (2): we know that y is a positive integer 1, 2, 3, ... so 1 - y must be less than or equal to 0.
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Re: M02-25  [#permalink]

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New post 28 Nov 2014, 05:30
positive integer means we should not consider zero??
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New post 28 Nov 2014, 05:45
sudd1 wrote:
positive integer means we should not consider zero??


Yes, positive integers can't be zero. It will start from 1
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New post 28 Nov 2014, 07:25
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New post 20 Apr 2015, 22:00
Bunuel, from what I understand about absolute values, the number/expression within the absolute number sign could either be positive or negative (just like the way the square root of a number could either be positive or negative), so I didn't understand how the least value in the LHS is zero. Please explain? Thanks
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New post 21 Apr 2015, 04:25
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Kigunda wrote:
Bunuel, from what I understand about absolute values, the number/expression within the absolute number sign could either be positive or negative (just like the way the square root of a number could either be positive or negative), so I didn't understand how the least value in the LHS is zero. Please explain? Thanks


I think this is explained here: m02-183573.html#p1448632

The point is that |some expression| >= 0. So, when we have |some expression| = x, it means that x must be more than or equal to 0 too.

Also, the square root of a number cannot be negative \(\sqrt{4}=2\), not +/-2.

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).

Theory on Number Properties: math-number-theory-88376.html
Tips on Numper Properties: number-properties-tips-and-hints-174996.html

All DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
All PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59

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Re: M02-25  [#permalink]

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New post 16 Oct 2015, 07:18
I think this is a high-quality question and I agree with explanation. This question is the same as M28-58
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New post 20 Nov 2015, 18:38
Can someone help me understand this part in detail

2)x+y−3=|1−y|. Since y is a positive integer, then 1−y≤0, thus |1−y|=−(1−y).

why we are not considering the |1−y|=(1−y) ??
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New post 21 Nov 2015, 01:46
sajib2126 wrote:
Can someone help me understand this part in detail

2)x+y−3=|1−y|. Since y is a positive integer, then 1−y≤0, thus |1−y|=−(1−y).

why we are not considering the |1−y|=(1−y) ??


Let me ask you if y is a positive integer what are the possible values of 1 - y ?
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Re: M02-25  [#permalink]

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New post 23 Dec 2015, 08:43
Bunuel wrote:
If \(x\) and \(y\) are positive integers, is \(x\) a prime number?


(1) \(|x - 2| \lt 2 - y\)

(2) \(x + y - 3 = |1-y|\)



I've solved this problem in a little different way.

1) |x - 2| < 2 - y It means
-(2 - y) < x - 2 < 2 - y =>
y - 2 < x - 2 < 2 - y =>
y < x < 4 - y

Try pick up numbers. It's said that x and y are positive integers.
Let's say y is 1, then x could be 2 or 3.
y can't be 2, 3 or any other integer, because it violates the inequality.
In both cases, whether x is 2 or 3, x is a prime number. Therefore, first statement is sufficient.

2) x + y - 3 = |1 - y| It means
x + y - 3 = 1 - y or x + y - 3 = -(1 - y), x + y - 3 = y - 1
x = 4 - 2y or x = 2 . 2 is a prime number. Therefore, second statement is sufficient too.

Thus, correct answer is D: both statements are sufficient.
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Re: M02-25  [#permalink]

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New post 17 Jul 2016, 09:37
I think this the explanation isn't clear enough, please elaborate. for option 1 my logic is
since x is a positive integer, lx-2l>0, therefore
x-2<2-y
x+y<4
so, X can be 1 or 2, cannot determine
can someone correct me, if my logic is wrong?
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Re: M02-25  [#permalink]

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New post 17 Jul 2016, 09:40
r19 wrote:
I think this the explanation isn't clear enough, please elaborate. for option 1 my logic is
since x is a positive integer, lx-2l>0, therefore
x-2<2-y
x+y<4
so, X can be 1 or 2, cannot determine
can someone correct me, if my logic is wrong?


(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 \(0 \lt 2 - y\), thus \(y \lt 2\).
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Re: M02-25  [#permalink]

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New post 04 Sep 2016, 03:28
then 1−y≤01−y≤0, thus |1−y|=−(1−y)|1−y|=−(1−y)


How is that |1−y|=−(1−y)|1−y|=−(1−y) always?

How about the case when y =1 , shouldn't it be (1-y) only then??
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New post 04 Sep 2016, 04:06
victoraditya wrote:
then 1−y≤01−y≤0, thus |1−y|=−(1−y)|1−y|=−(1−y)


How is that |1−y|=−(1−y)|1−y|=−(1−y) always?

How about the case when y =1 , shouldn't it be (1-y) only then??


Why? What would be -(1-y) and |1-y| for y=1?
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Re: M02-25  [#permalink]

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New post 04 Sep 2016, 05:12
Hi Bunnel,
got it.. so is it that for <=0 inequalities, we can always take the negative?

Thanks.
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New post 12 Dec 2016, 19:04
Bunuel wrote:
Official Solution:

If x and y are positive integers, is x a prime number?

(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 \(0 \lt 2 - y\), thus \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y-2 \le{0}\) 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 \(1-y \le {0}\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.


Answer: D



Thanks for the solution. But, i can`t understand one point. In statement 1, you mentioned if y is more than or equal to 2, then y−2≤0y−2≤0 and it cannot be greater than |x−2||x−2|. Could you please explain this point in more detail?

Thanks in advance.
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New post 13 Dec 2016, 00:55
jahidhassan wrote:
Bunuel wrote:
Official Solution:

If x and y are positive integers, is x a prime number?

(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 \(0 \lt 2 - y\), thus \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y-2 \le{0}\) 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 \(1-y \le {0}\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.


Answer: D



Thanks for the solution. But, i can`t understand one point. In statement 1, you mentioned if y is more than or equal to 2, then y−2≤0y−2≤0 and it cannot be greater than |x−2||x−2|. Could you please explain this point in more detail?

Thanks in advance.


If y is greater than 2 then the right hand side of the inequality (2-y) becomes negative. The left hand side if the inequality is an absolute value (|x - 2|) which cannot be negative, thus the inequality (\(|x - 2| \lt 2 - y\)) will not hold true. Please re-read the discussion above.
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Re: M02-25 &nbs [#permalink] 13 Dec 2016, 00:55

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