Confused!

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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

[[email protected]]

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

[theCodeToGMAT]

a^2 + b^2 = c^2

a*a = (c-b)(c+b)

To find: (c-b)^2 = c^2 + b^2 - 2bc

Statement 1:
"a" is prime
Since, the numbers are positive.. then
(c-b) must be "1"
(c+b) = a*a
SUFFICIENT

Statement 2:
b^2 is multiple of "4"
b*b = 4x_
We done have any information about the values of "a" or "c"
INSUFFICIENT

Answer [spoiler]{A}[/spoiler]

[[email protected]]

Rahul why did u take c-1 as 1?

a^2 + b^2 = c^2

a*a = (c-b)(c+b)

To find: (c-b)^2 = c^2 + b^2 - 2bc

Statement 1:
"a" is prime
Since, the numbers are positive.. then
(c-b) must be "1"
(c+b) = a*a
SUFFICIENT

Statement 2:
b^2 is multiple of "4"
b*b = 4x_
We done have any information about the values of "a" or "c"
INSUFFICIENT

Answer [spoiler]{A}[/spoiler]

[[email protected]]

Rich can we solve this other than by this method of remembering this rare rule?

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

[theCodeToGMAT]

Rahul why did u take c-1 as 1?

a*a = (c-b)(c+b)

Since "a" is a prime number .. also we know that a,b and c are different positive integers..

So, it is not possible that (c-b) & (c+b) will yield same result "a" .. that means either (c-b) is a*a or (c+b) is a*a and the other would be "1".

So, (1) * (a*a) = (c-b) * (c+b)

(c+b) cannot be "1" as a,b,c are distinct positive integers. that means (c-b) is "1"

(c-b)^2 = (1)^2 = 1

Hope it's better now.

[stevecultt]

Excellent explanation!

Rahul why did u take c-1 as 1?

a*a = (c-b)(c+b)

Since "a" is a prime number .. also we know that a,b and c are different positive integers..

So, it is not possible that (c-b) & (c+b) will yield same result "a" .. that means either (c-b) is a*a or (c+b) is a*a and the other would be "1".

So, (1) * (a*a) = (c-b) * (c+b)

(c+b) cannot be "1" as a,b,c are distinct positive integers. that means (c-b) is "1"

(c-b)^2 = (1)^2 = 1

Hope it's better now.

[[email protected]]

Hi shibsriz,

Yes, we can solve this DS question by gathering enough evidence of a pattern. My explanation TESTed Values based on the given information. As a general rule, if you can come up with 3 examples that lead to a consistent result, then the Fact is likely Sufficient. This requires detailed work on your part AND you have to be thorough in your thinking.

GMAT assassins aren't born, they're made,
Rich