Re: [Math4u] A Combinatorial Problem

I have come up with 780 numbers too.

 

Using the product and sum rules carefully:

 

Assume the digits are UZHTN= N*1+T*10+H*100+Z*1000+U*10000 (i.e. N=ones, T=tens, H=hundreds, Z=thousands, and U= tens of thousands)

 

5 possibilities for N (0, 1, 2, 3, 4) (the 9 is possible too, but will br treated exceptionally).

again, 5 possibilties for T, 5 possibilities for H, 5 possibilities for Z, and one possibility for U)

This gives 1*5*5*5*5=5^4.

 

When N=9, again, we have 5 possibilties for T (with exceptional possibilty that T=9), 5 possibilities for H, 5 possibilities for Z, and one possibility for U. This gives 1*5*5*5*1= 5^3.

 

When N=9 and T=9, once again, we have 5 possibilities for H (with exceptional possibilty that H=9), 5 possibilities for Z, and one possibility for U. This gives 1*5*5*1*1= 5^2.

 

When N=9, T=9 and H=9, again, we have 5 possibilities for Z (If the 20000 was included, then we would have the exceptional possibilty that Z=9), and one possibility for U. This gives 1*5*1*1*1=5^1.

 

(if the 20000 is included, then we would have in addition: 1*1*1*1*1=5^0)

 

Total= 5^4+5^3+5^2+5^1= 780.

 

This is the type of discussion we want, like, and need.

I hate other very arguable very minor time wasting discussions.

 

Mosaad Alabdullatif 

"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

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