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If x and y are positive integers, is x a prime number? [M28-58]

(1) |x−2|<2−y.

(2) x+y−3=|1−y|

"If x and y are positive integers" implies x and y are 1/2/3/4... etc

(1) |x−2|<2−y Minimum value of y is 1 which means maximum value of right hand side is 2-1 = 1. An absolute value cannot be negative so left hand side must be 0 only (to be less than 1 but be an integer)

|x−2| = 0 means x = 2 (prime number) This statement alone is sufficient.

(2) x+y−3=|1−y| x = |1−y| - y + 3

If y = 1, x = 0 - 1 + 3 = 2 If y is greater than 1, (1-y) will be negative so |1−y| = -(1 - y) = y - 1 So x = y - 1 - y + 3 = 2

So for all cases, x must be 2. This statement alone is sufficient.

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

(1) |x - 2| < 2 - y . The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus 0 < 2 - y, thus y < 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: \(|x - 2| < 1\), which implies that \(-1 < x-2 < 1\), or \(1 < x < 3\), 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.

Re: If x and y are positive integers, is x a prime number? [#permalink]

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05 Jul 2014, 06:23

2

This post received KUDOS

This is one of the classic question. (1). |x−2|<2−y. The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus ,2-y>0 thus y<2 . Next, since given that is a positive integer, then y=1.

So, we have that:|x-2|<1 , which implies that ,-1<x-2>1 thus x=2. Sufficient.

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

we can write this in two form , considering positive & negative

(a) x+y-3 = 1-y x=1-y-y+3 => x= 1-2y+3 => x= 2(2-y) Since x is positive integer so y can be greater than 2 so y has to be 1. So x=2

(b) x+y-3 = y-1 x=2. So by both ways X=2. Each statement is sufficient
_________________

Re: If x and y are positive integers, is x a prime number? [#permalink]

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21 Aug 2014, 22:23

Hi

Isn't 0 a positive interger. If isn't whenever the question mentions positive or negative intergers we do not take 0...right

what about non-negative integers which include...0.1..2......

VeritasPrepKarishma wrote:

bekerman wrote:

If x and y are positive integers, is x a prime number? [M28-58]

(1) |x−2|<2−y.

(2) x+y−3=|1−y|

"If x and y are positive integers" implies x and y are 1/2/3/4... etc

(1) |x−2|<2−y Minimum value of y is 1 which means maximum value of right hand side is 2-1 = 1. An absolute value cannot be negative so left hand side must be 0 only (to be less than 1 but be an integer)

|x−2| = 0 means x = 2 (prime number) This statement alone is sufficient.

(2) x+y−3=|1−y| x = |1−y| - y + 3

If y = 1, x = 0 - 1 + 3 = 2 If y is greater than 1, (1-y) will be negative so |1−y| = -(1 - y) = y - 1 So x = y - 1 - y + 3 = 2

So for all cases, x must be 2. This statement alone is sufficient.

Isn't 0 a positive interger. If isn't whenever the question mentions positive or negative intergers we do not take 0...right

what about non-negative integers which include...0.1..2......

VeritasPrepKarishma wrote:

bekerman wrote:

If x and y are positive integers, is x a prime number? [M28-58]

(1) |x−2|<2−y.

(2) x+y−3=|1−y|

"If x and y are positive integers" implies x and y are 1/2/3/4... etc

(1) |x−2|<2−y Minimum value of y is 1 which means maximum value of right hand side is 2-1 = 1. An absolute value cannot be negative so left hand side must be 0 only (to be less than 1 but be an integer)

|x−2| = 0 means x = 2 (prime number) This statement alone is sufficient.

(2) x+y−3=|1−y| x = |1−y| - y + 3

If y = 1, x = 0 - 1 + 3 = 2 If y is greater than 1, (1-y) will be negative so |1−y| = -(1 - y) = y - 1 So x = y - 1 - y + 3 = 2

So for all cases, x must be 2. This statement alone is sufficient.

Answer (D)

0 is neither positive nor negative integer (the only one of this kind).

Re: If x and y are positive integers, is x a prime number? [#permalink]

Show Tags

09 Nov 2017, 17:27

VeritasPrepKarishma wrote:

bekerman wrote:

If x and y are positive integers, is x a prime number? [M28-58]

(1) |x−2|<2−y.

(2) x+y−3=|1−y|

"If x and y are positive integers" implies x and y are 1/2/3/4... etc

(1) |x−2|<2−y Minimum value of y is 1 which means maximum value of right hand side is 2-1 = 1. An absolute value cannot be negative so left hand side must be 0 only (to be less than 1 but be an integer)

|x−2| = 0 means x = 2 (prime number) This statement alone is sufficient.

(2) x+y−3=|1−y| x = |1−y| - y + 3

If y = 1, x = 0 - 1 + 3 = 2 If y is greater than 1, (1-y) will be negative so |1−y| = -(1 - y) = y - 1 So x = y - 1 - y + 3 = 2

So for all cases, x must be 2. This statement alone is sufficient.

If x and y are positive integers, is x a prime number? [M28-58]

(1) |x−2|<2−y.

(2) x+y−3=|1−y|

"If x and y are positive integers" implies x and y are 1/2/3/4... etc

(1) |x−2|<2−y Minimum value of y is 1 which means maximum value of right hand side is 2-1 = 1. An absolute value cannot be negative so left hand side must be 0 only (to be less than 1 but be an integer)

|x−2| = 0 means x = 2 (prime number) This statement alone is sufficient.

(2) x+y−3=|1−y| x = |1−y| - y + 3

If y = 1, x = 0 - 1 + 3 = 2 If y is greater than 1, (1-y) will be negative so |1−y| = -(1 - y) = y - 1 So x = y - 1 - y + 3 = 2

So for all cases, x must be 2. This statement alone is sufficient.

If x and y are positive integers, is x a prime number? [M28-58]

(1) |x−2|<2−y.

(2) x+y−3=|1−y|

"If x and y are positive integers" implies x and y are 1/2/3/4... etc

(1) |x−2|<2−y Minimum value of y is 1 which means maximum value of right hand side is 2-1 = 1. An absolute value cannot be negative so left hand side must be 0 only (to be less than 1 but be an integer)

|x−2| = 0 means x = 2 (prime number) This statement alone is sufficient.

(2) x+y−3=|1−y| x = |1−y| - y + 3

If y = 1, x = 0 - 1 + 3 = 2 If y is greater than 1, (1-y) will be negative so |1−y| = -(1 - y) = y - 1 So x = y - 1 - y + 3 = 2

So for all cases, x must be 2. This statement alone is sufficient.

Re: If x and y are positive integers, is x a prime number? [#permalink]

Show Tags

09 Nov 2017, 22:58

VeritasPrepKarishma wrote:

soodia wrote:

VeritasPrepKarishma wrote:

(1) |x−2|<2−y.

(2) x+y−3=|1−y|

"If x and y are positive integers" implies x and y are 1/2/3/4... etc

(1) |x−2|<2−y Minimum value of y is 1 which means maximum value of right hand side is 2-1 = 1. An absolute value cannot be negative so left hand side must be 0 only (to be less than 1 but be an integer)

|x−2| = 0 means x = 2 (prime number) This statement alone is sufficient.

(2) x+y−3=|1−y| x = |1−y| - y + 3

If y = 1, x = 0 - 1 + 3 = 2 If y is greater than 1, (1-y) will be negative so |1−y| = -(1 - y) = y - 1 So x = y - 1 - y + 3 = 2

So for all cases, x must be 2. This statement alone is sufficient.

Re: If x and y are positive integers, is x a prime number? [#permalink]

Show Tags

10 Nov 2017, 11:21

From statement 1 : |x−2|<2−y [u][/u] HERE WE HAVE 2 CASES CASE 1 : => X-2< 2-Y FOR X≥2 => X+Y< 4 ONLY POSSIBLE WHEN x = 2 (PRIME) CASE 2 : OR 2-X < 2-Y FOR X<2 => X-Y>0 (ONLY POSSIBLE WHEN X IS 1 AND Y IS 0 BUT AS PER THE QUESTION X AND Y MUST BE POSITIVE INTEGERS AND 0 IS NOT A ONE ) HENCE DISCARD CASE 2 THUS FROM STATEMENT 1 WE HAVE X=2 (PRIME) SUFFICIENT FROM STATEMENT 2 WE HAVE 2 CASES HERE CASE 1 : X+Y-3 = 1-Y FOR Y ≤1 => X+2Y = 4 THIS GIVES ONLY POSSIBLE VALUE FOR X AS 2 (PRIME) CASE 2 : X+Y-3= Y-1 FOR Y>1 THIS GIVES AGAIN X=2 (PRIME) HENCE BOTH THE CASES GIVE X= 2 WHICH IS PRIME . HENCE STATEMENT 2 IS ALSO SUFFICIENT HENCE ANSWER IS D

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

(1) |x - 2| < 2 - y . The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus 0 < 2 - y, thus y < 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: \(|x - 2| < 1\), which implies that \(-1 < x-2 < 1\), or \(1 < x < 3\), 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.

Answer: D.

Hi Bunuel,

I have a question in statement 1 :

We know X is positive so we open the MOD for X>0

X-2 < 2-y

X < 4-y

X+y < 4

As we know 0 is niether +ve nor -ve, we are left out with values for X and y to be 1,2,3

Why do we have to only consider x=2 here?

Why can't the values be x=1, y=2 or x=1,y=1? They also satisfy the equation we got from statement 1

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

(1) |x - 2| < 2 - y . The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus 0 < 2 - y, thus y < 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: \(|x - 2| < 1\), which implies that \(-1 < x-2 < 1\), or \(1 < x < 3\), 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.

Answer: D.

Hi Bunuel,

I have a question in statement 1 :

We know X is positive so we open the MOD for X>0

X-2 < 2-y

X < 4-y

X+y < 4

As we know 0 is niether +ve nor -ve, we are left out with values for X and y to be 1,2,3

Why do we have to only consider x=2 here?

Why can't the values be x=1, y=2 or x=1,y=1? They also satisfy the equation we got from statement 1

1. We are told that x is positive not x - 2, so you certainly cannot say that |x - 2| = x - 2 only because x is positive. For example, if x were 1, then |x - 2| = -(x - 2) 2. Neither x = 1 and y = 2 nor x = 1 and y = 1 satisfy |x - 2| < 2 - y.

The question above is not that easy. You should be absolutely clear with fundamentals before attempting such questions: