brother how do u get the digits of number ???u use % and / rite??
On Fri, Aug 14, 2009 at 7:28 PM, Yogesh Aggarwal < yogesh.aggarwa...@gmail.com> wrote: > (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 MORE 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 7:25 PM, Yogesh Aggarwal < > yogesh.aggarwa...@gmail.com> wrote: > >> (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 >> > > > > -- > 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 -~----------~----~----~----~------~----~------~--~---
