(CORRECTED ALGO.) We can do like dis... - add all d digits of the no. - if the result is MORE than 10, add all the digits of the result again. - continue step2 if the result is still less than 10 - if the result is either 0, 3 , 6 or 9 den the no. is divisible by 3.
example 1 : num = 12345 sum1 = 15 (sum > 10) sum2 = 6 since sum < 10, we stop here and since final sum = 6 ....so the given no. is divisible by 3 On Fri, Aug 14, 2009 at 7:20 PM, santhosh venkat < santhoshvenkat1...@gmail.com> wrote: > given a number n > u can get the quotient when it is divided by 4 using right shift 2 times > like n >> 2 this ll give u quotient(q) > u can get the remainder by subtracting 4 * q from n which will give the > remainder when divided by 4 > > by doing this u ll express n as n = 4q + r = 3q + (q+r) > in this already 3q which is divisible by 3 .. u can apply the same logic > recursively to q+r and return the remainder obtained for q+r.. > > > On Fri, Aug 14, 2009 at 7:11 PM, Yogesh Aggarwal < > yogesh.aggarwa...@gmail.com> wrote: > >> @arun : we are not supposed to use / operator. but in ur algo u r using / >> or % has to be used to check wether the diff is divisible by 3. >> We can do like dis... >> - add all d digits of the no. >> - if the result is less than 10, add all the digits of the result again. >> - continue step2 if the result is still less than 10 >> - if the result is either 0, 3 , 6 or 9 den the no. is divisible by 3. >> >> example 1 : >> num = 12345 >> sum1 = 15 (sum > 10) >> sum2 = 6 >> since sum < 10, we stop here and since final sum = 6 ....so the given no. >> is divisible by 3 >> >> >> On Fri, Aug 14, 2009 at 3:09 PM, Arun N <arunn3...@gmail.com> wrote: >> >>> take an number find its binary >>> add all odd bits and even bits seperately >>> now check if the difference is divisible by 3 >>> if yes it is >>> say 6 110 -----> 1+0 - 1 =0 >>> 9 1001 -----> 1+0 - 0+1 = 0 >>> 12 1100 ------> 1+0 - 1+0 = 0 >>> Arun, >>> >>> On Fri, Aug 14, 2009 at 1:15 PM, richa gupta <richa.cs...@gmail.com>wrote: >>> >>>> >>>> can we check the divisibility of a given number by 3 withoutusing >>>> operators like '/' or '%'. >>>> I want the efficient solution to this problem .. >>>> >>>> can someone help ?? >>>> -- >>>> Richa Gupta >>>> (IT-BHU,India) >>>> >>>> >>>> >>> >>> >>> -- >>> Potential is not what U have, its what U think U have!!! >>> It is better to worn out than rust. >>> >>> >>> >>> >>> >> >> >> -- >> Best Wishes & Regards >> Thank You >> Yogesh Aggarwal >> B.Tech(IT), >> University School of Information Technology >> GGS Indraprastha University >> Delhi >> mailto: yogesh.aggarwa...@gmail.com >> #9990956582 >> >> >> > > > > -- Best Wishes & Regards Thank You Yogesh Aggarwal B.Tech(IT), University School of Information Technology GGS Indraprastha University Delhi mailto: yogesh.aggarwa...@gmail.com #9990956582 --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---
