If a,b,c are different positive integers,and a^2+b^2 = c^2, then what is the value of (c-b)^2?
I.a is a prime.
II.b^2 is a multiple of 4.
Ans-A
I.a is a prime.
II.b^2 is a multiple of 4.
Ans-A
Hi shibsriz,
This DS question is based on a remarkably RARE rule about the Pythagorean Theorem which you're not likely to see on the GMAT.
We're told that A, B and C are different positive integers and A^2 + B^2 = C^2. We're asked for the value of (C-B)^2?
Fact 1: A is PRIME. This is an interesting "restriction"; let's TEST VALUES
A = 3
B = 4
C = 5
Here, (5-4)^2 = 1
A = 5
B = 12
C = 13
Here, (13-12)^2 = 1
A = 7
B = 24
C = 25
Here, (25-24)^2 = 1
This is a consistent result, so Fact 1 IS SUFFICIENT. The rare rule that I mentioned earlier is that IF A, B and C are all different integers AND A (or B) is a prime, then the other two numbers will differ by 1. So (C-B)^2 will always = 1
Fact 2: B^2 = multiple of 4
A = 3
B = 4
C = 5
Here, (5-4)^2 = 1
A = 6
B = 8
C = 10
Here, (10-8)^2 = 4
Fact 2 is INSUFFICIENT.
Final Answer: A
GMAT assassins aren't born, they're made,
Rich
