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
If a and b are positive integers divisible by 6
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ii) a=3b. Hence the two nos are b and 3b. Their GCD will be b. But b can be any no divisible by 6 (i.e. 6, 12, 18, etc) ==> insufficient.
i) a=2b+6
Hence the two nos are 2b+6, b. Pick nos for b i.e. 6, 12, 30 ..
we get a= 18, b=6 GCD = 6
a=30, b=12 GCD = 6
a= 66, b= 30 GCD = 6
and so on..
==> SUFFICIENT
Ans: a (statement 1 is sufficient).
Please let us know if this is the correct ans.
i) a=2b+6
Hence the two nos are 2b+6, b. Pick nos for b i.e. 6, 12, 30 ..
we get a= 18, b=6 GCD = 6
a=30, b=12 GCD = 6
a= 66, b= 30 GCD = 6
and so on..
==> SUFFICIENT
Ans: a (statement 1 is sufficient).
Please let us know if this is the correct ans.
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a = 6x
b = 6y
for 6 to be the HCF, x and y should be mutually prime. (no common factor among x and y)
statement 1 : a = 2b + 6
6x = 12y + 6
x = 2y+1
check for y = 1,2,3..... x=3,5,7 Suff.
Statement 2:
a = 3b
x=3y
x and y are mutually prime only in one case, (1,3) so 6 can / can not be HCF. Insuff
Answer A.
b = 6y
for 6 to be the HCF, x and y should be mutually prime. (no common factor among x and y)
statement 1 : a = 2b + 6
6x = 12y + 6
x = 2y+1
check for y = 1,2,3..... x=3,5,7 Suff.
Statement 2:
a = 3b
x=3y
x and y are mutually prime only in one case, (1,3) so 6 can / can not be HCF. Insuff
Answer A.
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