yes u r wrong.. 111111111 is nt divisible by 23 On Sat, Aug 6, 2011 at 9:15 AM, sumit <[email protected]> wrote:
> 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 <[email protected]> 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 <[email protected]> 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 [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. > > -- 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.
