LiewCF.com

I was told by a friend of mine that this is a question from a branch of math called "discrete math". It's mostly a sub topic of computer science. It's quite a though question.

I guess this if for there programming purpose. Maybe for their grabber or something else.

You are a genius to be able to solve this!

One of the ways I tried to approach it was to think of it this way: let's say there are s stairs and you are given n times to jump up the stairs. Each time you can choose between jumping 0 stairs to 9 stairs, but the first time you have to jump at least one stair. How many ways you can choose to jump up the stairs?

I have spent a whole day on this and all I could get was a solution for n

which is less than 18. For greater than 18 it gets too complicated for my methods. Going to give up... Sigh. :P

Their question looks pretty simply actually. A good add math student should be able to answer that question..Too bad I'm not..I've finished school long time ago..so whatever I learnt bout the subject is like ingat-ingat lupa..:D

It's not that simple... The problem is you have to find a general formula for any n and s, so it's not as simple as it seems. :( For example, Fermat's Last Theorem also looked quite simple but it took world's best mathematicians 300 years to figure it out..

n*s

this is the answer prove me wrong by giving me a solution and i will give u 5 rapidshare premium accounts...i promise!!!!

Example:

for n=2, s=5 all combinations are: 14, 23, 32, 41 and 50

so answer is: There are 5 combinations in total.


Your solution has to be general formula (and not like: if n=XXX and S=YYY there are WWW combinations). e.g. permutations, variation, combinations (with and without repeating) you still have to denote and explain all symbols (functions) you use.

i don't even understand the question coz even my english is quite good but not 'mathematician english'(is this correct?)...LOL!

im gona go with infinity

They closed the competition already, and nobody solved it. :(

I get the answer 0.

For n=2, s=5 all combinations are:

14

23

32

41 and

50

Answer : There are 5 combinations in total.
& the formula should be
{s+n-1-[(s+8)/9]+1}

I think the formula is:

S= n*(n+1)/2

Am I right?

Is too late?

i guess you all misunderstood the question (or it's me) coz it says "how many numbers of"?
lets say for example, n=5 digits and s=21 and the number is 55551. that means there are 4 5s or the number is 54651. there are 2 5s. i think ppl must concentrate on this.
not like n=3 s=5 then S={104,113,122,131,140,203,212,221,230,302,311,320,401,410}

The final answer (certainly true):
Let k be the number of integers of n digits which sum of all digits is s
If (s=0), then (if n=1, then k=1; else k=0);
else k = (n+s-2)C(s-1)
where nCr = n!/((n-r)!*r!)

number of digit is = s/n

am i right?

DarkAngel, your answer is fabulous! Do you mind sharing your working? (Don't worry about people stealing the answer because the competition is already closed)

I just tested your solution - it works well for low numbers, but it doesn't work for certain cases... For example, let's say s = 26, and n = 3. So k=3 as the only possible numbers are 998, 989 and 899. However, your answer will give the answer of 27C25=351.

That was my dilemma too when I tried the question... the problem with this question is, you have to consider that each digit can only take the value of 1 to 9, not 10 and above. :(

@chang yang

How many numbers of n digits exist if the sum of all digits is s?

where n and s is the only required given plus all the digits

if s=26
there is posibilities of 1+2+3+4+5+2+3+6 = 26
therefore 12345236 = 8digits as n

i just wonder :D

wel i like to know if you awarded the guy
did he win hees free account

I like to think of it in this way:
The sum of all digits, i think there's 10 digits (0,1,2,3,4,5,6,7,8,9,)
So then the SUM of all digits would be 0+1+2+3+4+5+6+7+8+9 = 45
How many numbers of digits in the number 45? There's two digits. (4 and 5)
So my final answer would be:

There is 2 digits in the sum of all digits.
n=2
s=45

Stephan: I think you have misunderstood the question. When it says "if the sum of all digits is n", it doesn't mean n is the sum of 1, 2, 3, 4 ... , 8, 9. Instead, the question is as the following:

How many 2-digit numbers are there of which the sum of their digit is 5? (e.g. 23, 41, 14, 50, 32)
How many 3-digit numbers are there of which the sum of their digit is 9?
How many 4-digit numbers are there of which the sum of their digit is 6?

And the REAL question there is... for the questions like the 3 I just gave above, what is the the general formula to calculate the answer. In other words, how many n-digit numbers are out there of which the sum of digit is s?

The answer is : n

because the sum of all s digits is s*n

so the answer is n

i guess you all misunderstood the question (or it’s me) coz it says “how many numbers of”?
lets say for example, n=5 digits and s=21 and the number is 55551. that means there are 4 5s or the number is 54651. there are 2 5s. i think ppl must concentrate on this.
not like n=3 s=5 then S={104,113,122,131,140,203,212,221,230,302,311,320,401,410}

The answer is infinity .
let s=1 that is sum of :

10000000000000000000000000000000000000
10000000000000000000000000000000000000000000000000000000

regards

Well, I found a general formula, a recurring sequence (don't know if it's called that way in English...). It can be kinda hard to understand. If you want, I can show you an excel sheet showing what I mean with this.

Being y the number of integers of n digits with sum s:
y(n=1, 19) = 0
y(n, s

Sorry, seems this site doesn't accept characters used on tags, I'll put Imageshack link:

http://img179.imageshack.us/img179/4642/formula...

Simplified, it means, for example, that y(3,12) = y(2,3)+y(2,4)+y(2,5)+...+y(2,12). The result is 66, being the sum of 3+4+5+6+7+8+9+9+8+7. y(6,3) is only the sum of y(5,1)+y(5,2)+y(5,3), because y(5,0) and s smaller than 0 is always 0.

All results tested, and they were right. If you find some error, tell me ;)

if we have n digit .. and each digit between 0 & 9 ==>

we have (9 ^ n ) numbers

am i right

S= n*(n+1)/2

the answer to the question is very simple :--
S=(n/2)*(2a+(n-1)d)
where
s=sum of the numerical s
n=number of numerical
a=the first numerical
d=difference in between the first 2 numerical

17

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.

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

I'm just splitting my reply. sorry

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.

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

Lol, sorry again but i see the last part of my post got cut off. I changed the string abit so now there should be no prob.


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

It was a lengthy and great explanation, Ben, thanks a lot. Actually I got that far into the question, but the problem that I encountered, and I think what you have not considered, is: you can't put more than 9 balls into _any_ box. And that has made this solution of yours useful only for lower value numbers of s. So for questions such as s=26, n=3, the answer is obviously 3 (998, 989, 899) but your answer will turn it into 27C25 which is obviously not 3.

Please note that because of certain problematic strings all of my >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.

The top got cut off again, sorry


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.

Ok, this is the real first part, sorry about all these repeted posts.


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.

chang please send me your mail and I will send you my solution since some strings are cutting off parts of the reply.
thank you.

How many numbers of 'n' digits exist if the sum of all digits is 's'?
You can say a number as below

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'.

s/n if the number of digits is 4 (n) then the total number of digits has to be divisible by 4 or 4,8,12,16 etc and dividing this number (s) by (n) will give you the number. 16/4 = 4 numbers of 4 digits each.

huh? are you really gonna be using any of these methods. i think its a wast of my time.

the aswer is simple. Its: "none" . Because if sum all digits you only get an s so s is not a n so theres none

The sum of proper divisors gives various other kinds of numbers. Numbers where the sum is less than the number itself are called deficient, and where it is greater than the number, abundant. These terms, together with perfect itself, come from Greek numerology. A pair of numbers which are the sum of each other's proper divisors are called amicable, and larger cycles of numbers are called sociable. A positive integer such that every smaller positive integer is a sum of distinct divisors of it is a practical number.

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

clearly we have x1+x2+......xN = s
and all x are integers between 0 and 9 and also x1 cannot be zero.

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

Here's the latest RS accounts guys, tested and checked as of 21:48 16.02.2007
Please dont change passwords!

http://rapidshare.com/files/16802515/RS_Account...

latest archive of hacked rapidshare accounts just for anyone who wants piracy :
by :

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

it just works untill 28 mar 2007

http://rapidshare.com/files/16841447/Rapidshare...

enjoy that my friend :)

10^n-10^(n-1)

there are not working

please give me working premium accounts

want rapidshare, megaupload, megashare, filefactory, depositefiles account ??
no one can give it to you except me if anyone of you make 300 posts on my forum !! - www.dhoom.bigbig.com , here you can find everything to download if you register and give links which you know !!