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Re: If y is an integer, is y^3 divisible by 9 ? [#permalink]
29 Jan 2014, 02:07
From S1:y=4k y^3=64*k^3.Insufficient as K could take any value and accordingly y^3 could or could not be div by 9. From S2:y=6k y^3=216k^3.Regardless of value of k,y^3 will be div by 9 because 216 is div by 9. Sufficient. Ans.B
IMO B takind A alone, if y is divisible by 4 then y= 4n therefore , y^3 = (4n)^3 = 64n^3 . thus we cant say anything about the divisibility by 9 since it will not depend totally on the value of n INSUFFICIENT
taking B alone, if y is divisible by 6 then y= 6n therefore , y^3 = (6n)^3 = (2*3*n)^3 = (8* 9*3) n^3. thus we can say that it is divisible by 9. SUFFICIENT.