This looks quite simple. Every number ending in 3 follows a pattern.eg- 3 - 111 13 - 111111 23 - 111111111 etc we can find the reauired no. by : suppose input no. is 33 In every case leave the no at 1's place(least significant) i.e. 3, In 33 you will be left with 3(after removal of 3 at first place). Now ,3 *(rest of nos +1 ) is your answer (in case of 33 it is 3*(3+1) = 12 i.e 111111111111). for 103 it is 3*(10+1) = 33 1's.
Correct if I am wrong. On Aug 5, 4:33 pm, Manee <mani.ma...@gmail.com> wrote: > ADOBE asks the very basic C/C++ questions > > one of their toughest however was : > > every number ending in 3 has a multiple of the form "111...111" > > e.g 3 has 111 > 13 has 111111 > so on.. > > find the algo for finding the number for an input number ending in 3. > > On Aug 5, 2:33 pm, Agyat <jalsa.n.sa...@gmail.com> wrote: > > > > > > > > > hey, guys adobe is visiting our campus. So those who know questions > > that adobe asked in written or interview, please post here as it will > > be of great help (as adobe has visited some colleges already). > > Thank you in advance. -- 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?hl=en.
