[Math4u] Re: A Combinatorial Problem

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@gmx.net> 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@gmx.net> 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

---

Be a better friend, newshound, and know-it-all with Yahoo! Mobile. Try it now.

---

Looking for last minute shopping deals? Find them fast with Yahoo! Search. __._,_.___

Your email settings: Individual Email|Traditional
Change settings via the Web (Yahoo! ID required)
Change settings via email: Switch delivery to Daily Digest | Switch to Fully Featured
Visit Your Group | Yahoo! Groups Terms of Use | Unsubscribe

__,_._,___