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# M02-25

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:18
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95% (hard)

Question Stats:

33% (01:43) correct 67% (01:47) wrong based on 246 sessions

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

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

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

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Joined: 02 Sep 2009
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16 Sep 2014, 00:18
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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.

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27 Nov 2014, 12:44
[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?
Math Expert
Joined: 02 Sep 2009
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28 Nov 2014, 05:33
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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28 Nov 2014, 06:30
positive integer means we should not consider zero??
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28 Nov 2014, 06:45
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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28 Nov 2014, 08:25
sudd1 wrote:
positive integer means we should not consider zero??

Positive integers are integers which are to the right of 0 on the number line: 1, 2, 3, 4, ...
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20 Apr 2015, 23:00
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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21 Apr 2015, 05:25
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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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16 Oct 2015, 08:18
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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20 Nov 2015, 19:38
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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21 Nov 2015, 02:46
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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23 Dec 2015, 09:43
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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17 Jul 2016, 10:37
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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17 Jul 2016, 10:40
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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04 Sep 2016, 04:28
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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04 Sep 2016, 05:06
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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04 Sep 2016, 06:12
Hi Bunnel,
got it.. so is it that for <=0 inequalities, we can always take the negative?

Thanks.
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Joined: 23 Jun 2016
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12 Dec 2016, 20:04
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.

Thanks for the solution. But, i cant 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?

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13 Dec 2016, 01:55
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.

Thanks for the solution. But, i cant 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?

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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