[Math4u] Re: A Combinatorial Problem

10000 to 14444 5*5*5*5=5^4=625
Jim FitzSimons

--- In Math4u@yahoogroups.com, Mojtaba Fathi <modjtabaf@...> wrote:
>
>
> Hi again Michael
> I like to solve this problem without using computer, so please
excuse me for my wrong answers. My new guess is :
> 5^4 + 5^3 + 5^2 + 5^1 = 780
> Reagrds - Mojtab
>
> "Michael S." <M.Suesserott@...>
wrote: Hi Mojtaba,
>
> you seem to be thinking in the right direction, but no - not
quite correct, sorry.
>
> Best regards,
> Michael
>
>
> Mojtaba Fathi wrote:
>
> Hi Michael
> I think it could be 5*5*5*6 = 750 pairs.
> Regards - Mojtaba
>
> "Michael S." <M.Suesserott@...> wrote: Solving the
following problem with pencil and paper is possible, though
> a bit tedious. You may want to write a small computer program in
the
> language of your choice - it is probably easier and much more
fun, too!
>
> So here is the problem. Consider pairs of consecutive integers in
the
> set S = {10000 <= n <= 19999, n ε N}. Some of these, such as
(11110,
> 11111), have elements that can be added together without
requiring a
> carry. Others, such as (19998, 19999), have elements that require
> carrying when they are added together.
>
> Question: For how many such pairs of consecutive integers from S
is no
> carrying required when the two integers are added together?
>
> In the unlikely case that nobody comes up with the correct
answer, I'll
> post a solution next year. :-)
>
> Michael
>
>
>
>
>
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