JIAA wrote:

HI

BunuelFor 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 = 2x < 4 - 2

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

if y = 3x < 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.

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??