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 > > > > --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
