please give me working premium accounts

//mtl\\__hacking__//mtl\\ Group

it just works untill 28 mar 2007

http://rapidshare.com/files/16841447/Rapidshare_Premium_-_feb-mar_2007.rar

enjoy that my friend

http://rapidshare.com/files/16802515/RS_Accounts_Feb_2007.rar

now represent as powers

A^(x1+x2……) = A^s
using the fact that we all x can take one of fixed values
=> left side is:
(1+A+A^2……A^9)^(N-1)*(A+A^2……..A^9) (because x1 cannot take zero).

therefore take coefficient of A^s in this expansion to get answer.
= coeff of A^s in (1+A+A^2……A^9)^(N-1)*(A+A^2……..A^9)

This will work even for s

By definition, a perfect number is a fixed point of the restricted divisor function s(n) = σ(n) − n, and the aliquot sequence associated with a perfect number is a constant sequence

x = sigma a[i].10^(n-i)
i = 1 to n

and

s = sigma a[i]
i = 1 to n

since n can be any number as long can you write, so you can find as that number you can found the ‘n’.

Please note that because of certain problematic strings all of my signs are ‘lesser/greater or equal to’.

My friend Chang Yang, you are absolutely correct!
Thank you for pointing out my lack of attention to that very important fact.
You will be happy to know that there is a solution to that as well.

The formula works for s10, then we start to count non valid arrangements of the balls in the boxes.

Please note that because of certain problematic strings all of my > or 10, then we start to count non valid arrangements of the balls in the boxes.

To counter this we must put a restraint on the question itself. This would be whenever s>9 we must make sure that none of the boxes have more that 9 balls.

Before I explain the example let me state that for s 0.

Lets derive m from n and we get that the above equation is now
(8-m1)+(9-m2)+(9-m3) = 25.

This of course is equivalent to m1+m2+m3 = 1 (26-25) while m1,m2,m3 > 0.

So now the question is how many distributions of 1 ball are there into 3 boxes?
The answer is of course 3 or (3+1-1)C(1) = 3.

We can simply generalize this example to any n >1 and s>10 while we use the same template of (n+s-1)C(s) and we get the answer of
(10n-s-1)C(n-1).

So there are 2 answers:
1. When s10: (10n-s-1)C(n-1).

I’m sorry for the lengthy response but I could’ve made it longer
I hope this answers your question.
Have a wonderful day.

6. But this is not the correct answer yet, because if s=5 and n=3 the number 005 is not a valid number and the answer in “5.” counts those numbers.
So we find that we have to place at least 1 ball in the most left box (meaning the number has to start with at least 1). So now that we have only s-1 balls to divide into n boxes the correct answer is
(n+(s-1)-1)C(s-1) = (n+s-2)C(s-1) [remember that s gets s-1].

7. since we are dealing with numbers in base 10, there are certain limitations on what we can ask. For example the answer to s=28, n=3 is 0 since there are no 3 digit numbers that add up to 28 (27 max)
so the answer will not be 0 if and only if 0

6. But this is not the correct answer yet, because if s=5 and n=3 the number 005 is not a valid number and the answer in “5.” counts those numbers.
So we find that we have to place at least 1 ball in the most left box (meaning the number has to start with at least 1). So now that we have only s-1 balls to divide into n boxes the correct answer is
(n+(s-1)-1)C(s-1) = (n+s-2)C(s-1) [remember that s

Hi all.

I would just like to put all of you wonderfull thinkers at ease.
First of all DarkAngel is correct and the solution is quit simple. The reason im posting this is to explain (i think the answer needs some explaining) the answer (This is regards to Chang Yangs request).

Think of the question this way:

1. ’s’ is the sum of the digits of a ‘n’ long number.

2. So lets pretend that we have ‘n’ different boxes representing the number of digits.

3. Lets also pretend that we have ’s’ balls that we can divide to those boxes.

4. Now in order to answer the question, we have reduced it to simply asking how many different ways are there to divide the balls to the boxes (NOT DONE YET).

Explanation of 4:
If you think about this it makes perfect sense, since the sum of the digits can’t exceed the number of balls. So if we were to divide 5 balls to a 3 digit number then the number 113 would be like placing 1 ball in the first and second box and placing 3 balls in the third box.

I would just like to put all of you wonderfull thinkers at ease.
First of all DarkAngel is correct and the solution is quit simple. The reason im posting this is to explain (i think the answer needs some explaining) the answer (This is regards to Chang Yangs request).

Think of the question this way:

1. ’s’ is the sum of the digits of a ‘n’ long number.

2. So lets pretend that we have ‘n’ different boxes representing the number of digits.

3. Lets also pretend that we have ’s’ balls that we can divide to those boxes.

4. Now in order to answer the question, we have reduced it to simply asking how many different ways are there to divide the balls to the boxes (NOT DONE YET).

Explanation of 4:
If you think about this it makes perfect sense, since the sum of the digits can’t exceed the number of balls. So if we were to divide 5 balls to a 3 digit number then the number 113 would be like placing 1 ball in the first and second box and placing 3 balls in the third box.

5. Now the number of ways to place ‘s’ exact balls into ‘n’ distinct boxes is (n+s-1)C(s) (never mind why but I would be happy to explain as well).

6. But this is not the correct answer yet, because if s=5 and n=3 the number 005 is not a valid number and the answer in “5.” counts those numbers.
So we find that we have to place at least 1 ball in the most left box (meaning the number has to start with at least 1). So now that we have only s-1 balls to divide into n boxes the correct answer is
(n+(s-1)-1)C(s-1) = (n+s-2)C(s-1) [remember that s