zalim Posted May 13, 2005 Share Posted May 13, 2005 . If x and y are positive integers such that x=8y+12, what is the greatest common divisor of x and y? 1) x=12u, where u is an integer. 2) y=12z, where z is an integer. my answer is B...because in B...GCD comes to be 12 plz confirm Quote Link to comment Share on other sites More sharing options...
zalim Posted May 13, 2005 Author Share Posted May 13, 2005 Consider an extension to the problem....i.e. had x = 8y + 24...............then what would have been the answer....I want to know that how we can make sure that 12 is GCD in 2....is it because x - 8y = 12....therefore the GCD can not be greater than 12....so following the same reasoning can we infer that x - 8y = 24.....therefore GCD can not be greater than 24....is this some Rule? Quote Link to comment Share on other sites More sharing options...
arjmen Posted May 13, 2005 Share Posted May 13, 2005 . If x and y are positive integers such that x=8y+12, what is the greatest common divisor of x and y? 1) x=12u, where u is an integer. 2) y=12z, where z is an integer. my answer is B...because in B...GCD comes to be 12 plz confirm I: 12u = 8y +12 => 8y = 12*(u-1) Can have more than one possibility for HCF of x and y, due to the multiplicant in LHS above. We have, y = (3/2)*(u-1) and x = 12*u. Too many combinations possible. examples: If y = 3 => (u-1) = 2, and x = 12*3 = 36. HCF of x and y = 3 If y = 12 => (u-1) = 8, and x = 12*9 = 108. HCF of x and y = 12 Therefore I is insufficient II: y = 12*z and x = 12*(8z+1), z and (8z+1) cannot have common factors other than 1 Therfore the HCF of x and y = 12 Sufficient Ans B Quote Link to comment Share on other sites More sharing options...
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