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If a and b are positive integers divisible by 6

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by manu217 » Fri Jun 13, 2008 12:54 am
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.
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by durgesh79 » Fri Jun 13, 2008 1:08 am
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.
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by g_beatthegmat » Sun Jun 15, 2008 10:08 pm
nice approach to the problem durgesh79!
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