@sumit, these numbers containing all ones are not in binary representation.
They are in decimal system.

On Sat, Aug 6, 2011 at 9:51 AM, sahil gujral <[email protected]> wrote:

> 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.
>>
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-- 
Nikhil Gupta
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CSI, NSIT Students' Branch
NSIT, New Delhi, India

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