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Free RapidShare Account by Solving a Math Problem
Filed in: Web — October 31st, 2006
Here is your chance to get a free (and legal) RapidShare account! CMSZONE is giving away 30-days RapidShare.de or RapidShare.com account (worth 9.90 EUR) if you can solve the following math problem.
How many numbers of n digits exist if the sum of all digits is s?
Please refer to the static page for more detail information.
My two cents
Well, I am not good in math. This is not for me. 
Furthermore, I already have both RapidShare.de and a free RapidShare.com account.
By the way, are there many people wish to get a RapidShare account? If so, I might add RapidShare account as prize for future giveaway/contest. 
[Thanks, Jaynier]
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November 1st, 2006 at 2:15 am
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!
November 1st, 2006 at 2:18 pm
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
November 1st, 2006 at 2:19 pm
which is less than 18. For greater than 18 it gets too complicated for my methods. Going to give up… Sigh.
November 1st, 2006 at 3:04 pm
[...] Found via LiewCF [...]
November 1st, 2006 at 5:07 pm
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
November 1st, 2006 at 5:25 pm
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..
November 1st, 2006 at 6:53 pm
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!!!!
November 5th, 2006 at 12:49 pm
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.
November 7th, 2006 at 12:38 pm
i don’t even understand the question coz even my english is quite good but not ‘mathematician english’(is this correct?)…LOL!
November 8th, 2006 at 9:17 am
im gona go with infinity
November 9th, 2006 at 6:53 am
They closed the competition already, and nobody solved it.
November 9th, 2006 at 4:13 pm
I get the answer 0.
November 11th, 2006 at 5:28 am
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}
November 11th, 2006 at 8:41 am
I think the formula is:
S= n*(n+1)/2
Am I right?
Is too late?
November 11th, 2006 at 3:08 pm
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}
November 14th, 2006 at 11:48 am
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!)
November 14th, 2006 at 2:56 pm
number of digit is = s/n
am i right?
November 14th, 2006 at 4:01 pm
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.
November 15th, 2006 at 12:37 pm
@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
November 17th, 2006 at 10:19 am
wel i like to know if you awarded the guy
did he win hees free account
November 18th, 2006 at 11:36 am
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
November 19th, 2006 at 6:59 am
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?
November 21st, 2006 at 1:49 am
The answer is : n
because the sum of all s digits is s*n
so the answer is n
December 1st, 2006 at 1:39 pm
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}
December 1st, 2006 at 9:51 pm
The answer is infinity .
let s=1 that is sum of :
10000000000000000000000000000000000000
10000000000000000000000000000000000000000000000000000000
regards
December 5th, 2006 at 4:55 am
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
December 5th, 2006 at 5:06 am
Sorry, seems this site doesn’t accept characters used on tags, I’ll put Imageshack link:
http://img179.imageshack.us/img179/4642/formulafd7.jpg
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
December 14th, 2006 at 11:08 pm
if we have n digit .. and each digit between 0 & 9 ==>
we have (9 ^ n ) numbers
am i right
December 27th, 2006 at 11:32 pm
S= n*(n+1)/2
December 28th, 2006 at 11:53 pm
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
January 5th, 2007 at 3:44 pm
17
January 20th, 2007 at 12:55 am
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
January 20th, 2007 at 12:58 am
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.
January 20th, 2007 at 12:58 am
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
January 20th, 2007 at 1:01 am
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
January 20th, 2007 at 3:10 am
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.
January 20th, 2007 at 6:47 pm
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.
January 20th, 2007 at 6:49 pm
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.
January 20th, 2007 at 6:51 pm
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.
January 20th, 2007 at 6:54 pm
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.
January 24th, 2007 at 7:47 pm
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’.
February 7th, 2007 at 8:17 am
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.
February 8th, 2007 at 9:11 pm
huh? are you really gonna be using any of these methods. i think its a wast of my time.
February 10th, 2007 at 6:01 am
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
February 11th, 2007 at 1:56 pm
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
February 14th, 2007 at 2:57 pm
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
February 17th, 2007 at 6:15 am
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Please dont change passwords!
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February 17th, 2007 at 6:41 pm
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enjoy that my friend
February 20th, 2007 at 8:06 pm
10^n-10^(n-1)
February 22nd, 2007 at 7:50 pm
there are not working
please give me working premium accounts
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