If b is an integer, is root(a^2+b^2) an integer?
(1) a2 + b2 is an integer.
(2) a2 - 3b2 = 0
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Integer check
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patanjali.purpose
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root(a^2+b^2) an integer ==> possible only when a^2+b^2 is a perfect square ==> is (a^2+b^2) perfect sqGmatKiss wrote:If b is an integer, is root(a^2+b^2) an integer?
(1) a2 + b2 is an integer.
(2) a2 - 3b2 = 0
S1: a^2+b^2 is an interger but it will be perfect sq only when a & b gives rise to another number forming a pythagoras triplets. We can't say that for sure.
S2: a2 = 3b2 ==> a^2+b^2 = 4 b^2 = (2b)^2 ==> perfect sq
IMO B
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Statement 1: a² + b² is an integer.GmatKiss wrote:If b is an integer, is root(a^2+b^2) an integer?
(1) a2 + b2 is an integer.
(2) a² + b² = (√3)² + 1² = 4.
Let a=√3 and b=1 so that a² + b² = (√3)² + 1² = 4.
Then √(a²+b²) = √(√3² + 1²) = 2.
Let a=1 and b=1 so that a² + b² = 1² + 1² = 2.
Then √(a²+b²) = √(1² + 1²) = √2.
INSUFFICIENT.
Statement 2: a² - 3b² = 0.
Substituting a²=3b² into √(a²+b²):
√(a²+b²) = √(3b²+b²) = √(4b²) = 2b, which must be an integer since b is an integer.
SUFFICIENT.
The correct answer is B.
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saketk
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Stmt 1: a2 + b2 is an int.GmatKiss wrote:If b is an integer, is root(a^2+b^2) an integer?
(1) a2 + b2 is an integer.
(2) a2 - 3b2 = 0
now a can be anything say for example root (2) and let b = 4
this gives us:-- a2 + b2 = 2+16 = 18 (int)
but the root is not int.
Similarly we can choose any values of 'a' and 'b' which will give us a perfect square and the correponding root will be an integer.
Hence, stmt 1 is Not sufficient
Stmt 2: a2 = 3b2
this gives us a2 + b2 = 4b2
clearly this is a perfect square and the root of it will give us 2b ('b' being an integer) the number itself will be an integer.
Hence, stmt 2 is Sufficient.
IMO option B
