On Sat, Mar 5, 2011 at 5:37 PM, Gopi <kodaligopi...@gmail.com> wrote:
> n = 2^(b-1)

Shouldn't this be n = SUM (2^k) where k = 0 to b-1?

> each box should have x dollars where x is of from 2^y, y ranges from 0
> to (b-1).
>
> On Mar 5, 9:12 am, CodeTrooper <anujagrawa...@gmail.com> wrote:
>> I guess solution would be based on prime numbers.
>>
>> On Mar 5, 3:11 am, bittu <shashank7andr...@gmail.com> wrote:
>>
>>
>>
>>
>>
>>
>>
>> > “You have b boxes and n dollars. If I want any amount of money from 0
>> > to n dollars, you must be able to hand me 0 to b boxes so that I get
>> > exactly what I request.” The two questions were “What are the
>> > restrictions on b and n, and how is money distributed among the
>> > boxes?”
>>
>> > Thanks
>> > Shashank
>
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