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Rephrasing.

by vk_vinayak » Mon Aug 27, 2012 2:34 am
If a and b are both positive integers, is the square root of (b - a) also an integer?

(1) b > a + 21

(2) b = a(a + 1)

[spoiler]OA: B[/spoiler]
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by Anurag@Gurome » Mon Aug 27, 2012 3:55 am
considering option a alone
when b > a + 21
b - a > 21.
so b - a can be 22, 23, 24, 25 ..........
square root of 22 is not an integer, where as square root of 25 is an integer.
so, we can`t answer with option a alone.

considering option b alone
b = a(a+1) = a^2 + a
b - a = a^2
square root of b - a = square root of a^2 = a
a is given to be integer.

hence it is b
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by Brent@GMATPrepNow » Mon Aug 27, 2012 7:08 am
vk_vinayak wrote:If a and b are both positive integers, is the square root of (b - a) also an integer?

(1) b > a + 21
(2) b = a(a + 1)

[spoiler]OA: B[/spoiler]
Target question: Is sqrt(b - a) an integer?

Statement 1: b > a + 21
In other words, b - a > 21
This allows for many possible of values for b-a. Here are two cases:
case a: b-a = 25, in which case sqrt(b-a) is an integer
case a: b-a = 26, in which case sqrt(b-a) is not an integer
So statement 1 is NOT SUFFICIENT

Statement 2: b = a(a + 1)
Expand right side: b = a^2 + a
Subtract a from both sides: b-a = a^2
Take square root of both sides: sqrt(b-a) = sqrt(a^2)
If a is an integer, then sqrt(a^2) must be an integer, which means sqrt(b-a) is an integer
So statement 2 is SUFFICIENT and the answer is B

Cheers,
Brent
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