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
Could you explain Statement 1 as well? Thanks!
One approach is to plug in values.
Statement 1: a = 2b + 6
If b=6, then a = 2(6) + 6 = 18
The GCD of a=18 and b=6 is 6.
If b=12, then a = 2(12) + 6 = 30.
The GCD of a=30 and b=12 is 6.
One more combination if we want to be really, really safe.
If b=18, then a = 2(18) + 6 = 42.
The GCD of a=42 and b=18 is 6.
In every case, the GCD is 6.
SUFFICIENT.
Here's a proof.
Since a and b are each a multiple of 6, 6 is a factor of both a and b.
Thus, the GCD is at least 6.
The only question is whether a and b could have factors in common other than 6, in which case the GCD would be GREATER than 6.
Since b is a multiple of 6, let b = 6k, where k is an integer.
Statement 1: a = 2b + 6
Substituting b = 6k into a = 2b + 6, we get:
a = 2(6k) + 6
a = 6(2k + 1).
a has the following factors: 6 and 2k+1.
Since b = 6k, b has the following factors: 3 and 2k.
2k and 2k+1 are CONSECUTIVE integers.
Consecutive integers are COPRIMES: they share no factors other than 1.
Since 2k and 2k+1 have no factors in common other than 1, it is not possible for a and b to have a GCD greater than 6.
SUFFICIENT.
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