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17 Jun 2007, 01:35
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If a and b are positive integers divisible by 6, is 6 the greatest common divisor of a and b?
(1) a = 2b + 6
(2) a = 3b
Please explain
(1) a = 2b + 6
(2) a = 3b
Please explain
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17 Jun 2007, 05:55
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My answer is 'A'.
Stmt1: a =2b +6
the gcd between a and 2b may be 6, 12, 18....
But 2b+6 makes 'a' to be divisible by 6 not by the higher multiples of 6.
Hence the gcd of a and b is 6.
Suff
Stmt2: Clearly insuff.
Stmt1: a =2b +6
the gcd between a and 2b may be 6, 12, 18....
But 2b+6 makes 'a' to be divisible by 6 not by the higher multiples of 6.
Hence the gcd of a and b is 6.
Suff
Stmt2: Clearly insuff.
17 Jun 2007, 08:23
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A is sufficient
Bis not
Bis clearly not sufficient. Think like this - a = 3b,
b can be anything 6,12,18,24.. similarly GCD can be 6,12,18...
a = 2b+6 means that 6 will always be the GCD no matter what the value of b. g
Bis not
Bis clearly not sufficient. Think like this - a = 3b,
b can be anything 6,12,18,24.. similarly GCD can be 6,12,18...
a = 2b+6 means that 6 will always be the GCD no matter what the value of b. g
