well , I tried to solve it from Maths though half way through only :(
N = number required i.e. (10^k -1)/9 given n = 10x + 3 by eq (10x + 3) * m = N= (10^k - 1)/9 implies k = 2log 3 + log m + log(10x + 3) i.e. k > 1 + log n This gives the lowerbound on minimum number of 1 to be start with, but seeing the logarithmic eq, this simply seems not to add much value. can any one figure out for upper bound ? On Oct 12, 5:18 pm, anshu mishra <[email protected]> wrote: > @amol > I was not sure that for every number that has 3 in its unit place has one > multiple which has all one. So I used that is if the remainder is coming > that already appeared stop there coz it will make stuck in a loop. > for ex. remainders are > 1 3 19 23 37 1 3 19 .... that will repeat. > > but it in this case u can remove the set. Code will look more simpler. > > string all1Multiple(int x) > { > string s; > int r=1; > do > { > s += '1'; > r = r % x; > r = r * 10 + 1; > > } while(r != 1); > return s; > } -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
