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by [email protected] » Sun Nov 10, 2013 8:11 pm
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.


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Source: — Data Sufficiency |

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by [email protected] » Sun Nov 10, 2013 10:37 pm
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

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by theCodeToGMAT » Sun Nov 10, 2013 10:45 pm
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]
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by [email protected] » Mon Nov 11, 2013 2:25 am
Rahul why did u take c-1 as 1?

theCodeToGMAT wrote: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]
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by [email protected] » Mon Nov 11, 2013 2:26 am
Rich can we solve this other than by this method of remembering this rare rule?

[email protected] wrote: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.


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by theCodeToGMAT » Mon Nov 11, 2013 4:32 am
[email protected] wrote: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.
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by stevecultt » Mon Nov 11, 2013 5:57 am
Excellent explanation!
theCodeToGMAT wrote:
[email protected] wrote: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.
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by [email protected] » Tue Nov 12, 2013 12:28 am
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.

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