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M02-25

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Re: M02-25  [#permalink]

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New post 01 Apr 2019, 07:20
1
JIAA wrote:
Bunuel wrote:
JIAA wrote:
HI Bunuel

For S1:-

Since x & y > 0
then the expression |x - 2|< 2 - y can be simplified as follows:-
x -2 < 2 - y (if x>0 , then |x| = x)
x < 4 - y

Now, for y to be a positive integer y should be < 4 (i.e. y can be 1, 2 and 3)

Hence,

if y = 1,
x < 4 - y
x < 4 -1
x < 3 so x can be 1 or 2 ( 1 is NOT A PRIME & 2 IS A PRIME )

if y = 2

x < 4 - 2
x < 2 so x= 1 (1 is NOT A PRIME)

if y = 3

x < 4 - 3
x < 1 (this is not possible since x is a positive integer)



Can you please let me know where am i going wrong??

THANKS!


As you correctly wrote |x| = x, when x >= 0 (and |x| = -x, when x < 0).

So, |x - 2| = x - 2, when x - 2 >= 0, or which is the same, when x >= 2. In this case, so when x >= 2, x - 2 < 2 - y.
If y = 1, then x < 3 --> since we are looking for x >= 2, then x = 2.
If y = 2, then x < 2 --> since we are looking for x >= 2, then here we have no solution.
If y = 3, then x < 1 --> since we are looking for x >= 2, then here we have no solution.

Next, the second case would be when |x - 2| = -(x - 2) (|x| = -x, when x < 0). This would happen when x - 2 < 0, or which is the same, when x < 2. In this case, so when x < 2, -(x - 2) < 2 - y.
If y = 1, then x > 1 --> since we are looking for x < 2, then here we have no solution.
If y = 2, then x > 2 --> since we are looking for x > 2, then here we have no solution.
If y = 3, then x > 3 --> since we are looking for x > 2, then here we have no solution.

As you can see we have only one solution: x = 2 = prime.

Hope it's clear.



THANKS Bunuel

Just one more thing !


It's explicitly stated in Q stem that x is positive i.e. x> 0, shouldn't we consider second case ONLY when we don't know if x is positive or negative??

Why are we even considering second case here (i.e.|x| = -x, when x < 0)??



Would really appreciate your feedback!


Yes, x is positive integer but x - 2 (expression in modulus) could be negative (if x = 1), 0 (if x = 2) or positive (if x > 2).
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Re: M02-25  [#permalink]

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New post 15 Apr 2019, 11:52
Hi Bunuel can you please explain the highlighted blue part, I am not able to understand it.

I tried this problem as |x-2| < 2-y therefore there can be two answers 1) x-2<2-y => x<4-y
2) -x+2 < 2-y => -x<-y => x>y
Therefore it is insufficient
Bunuel wrote:


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.

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Re: M02-25  [#permalink]

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New post 14 May 2019, 11:41
sudd1 wrote:
positive integer means we should not consider zero??



Zero is neither positive or negative. Positive integers are to the right of zero on the number line.
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Re: M02-25  [#permalink]

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New post 14 May 2019, 11:57
rraman wrote:
Hi Bunuel can you please explain the highlighted blue part, I am not able to understand it.

I tried this problem as |x-2| < 2-y therefore there can be two answers 1) x-2<2-y => x<4-y
2) -x+2 < 2-y => -x<-y => x>y
Therefore it is insufficient
Bunuel wrote:


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.



I solved it the same way, but I got the correct answer. Here's how:

Statement (1):
|x−2|<2−y

This leads to two inequalities
First inequality:
x-2<2-y
x<4-y
x+y<4

Second inequality:
-x+2<2-y
-x<-y
x>y

Together this means that x+y<4 AND x>y, the only positive integers that make this work are if y=1 and x=2, therefore x is a prime number.

Statement 1 is sufficient.

Statement (2):
x+y−3=|1−y|

This leads to two equations
First equation:
x+y-3=1-y
x+2y-3=1
x+2y=4

Second equation:
x+y-3=-1+y
x-3=-1
x=2

Together this means that x+2y=4 AND x=2, therefore x is a prime number.

Statement 2 is sufficient.


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Re: M02-25  [#permalink]

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New post 25 May 2019, 22:17
Bunuel wrote:
ppnimkar wrote:
Hi Bunuel

Please let me know if my approach makes sense

Given: x, y -> +ve
x -> prime?

(1) |x-2| < 2-y
Opening up the Absolute Value:
when +ve
x-2 < 2-y
x+y < 4
OR when -ve
-x+2 < 2-y
-x<-y
x>y

Test Cases:
x=2
y=1
for x+y < 4 or = 3 [Given x, y > 0 & x>y, so total has to be 3]
and so x is prime

Sufficient

(2) x+y-3 = |1-y|
when +ve
x+y-3 = 1-y
x+2y = 4
Testing Cases:
x = 2, y = 1: Only option that works

OR when -ve
x+y-3 = -1+y
x=2 -> prime

Sufficient

Answer: D

Thanks
Prathamesh


I guess when you test values for (1), you consider x + y < 4 and x > y as simultaneous equations, and that is why you test only x = 2 and y = 1, whereas x + y < 4 and x > y are two different cases (true in certain conditions).


Hi Bunuel,
Can you explain why this way of solving is incorrect as I solved the question in a similar manner and arrived at the correct answer.
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Re: M02-25   [#permalink] 25 May 2019, 22:17

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