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**kappa_am****Member**- Registered: 2013-12-20
- Posts: 35

I have a number named X and N number of empty rooms. in how many ways I can fill this room to have X? all rooms have to be filled and all smaller number of N should be obtained by addin some of these numbers. and I know x is greater than N. is there any formula?

Example X=9 and N=4 there are the possibilities:

1,1,3,4; 1,1,4,3; 1,3,1,4; 1,3,4,1; 1,4,1,3; 1,4,3,1; 1,2,24; 2,2,1,4; 2,2,4,1; 4,2,2,1; 1,1,2,5;1,1,5,2; 5,1,1,2; 5,1,2,1; 2,1,1,5; 2,1,5,1; 2,5,1,1; 5,2,1,1,

thank you:)

*Last edited by kappa_am (2014-04-07 07:47:45)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,588

Hi kappa_am;

This is similar to the other question. Can we use the natural numbers up to X?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**kappa_am****Member**- Registered: 2013-12-20
- Posts: 35

I have modified the question. I have missed some information sorry.

all natural numbers can be used. gust there are 2 rules the numbers have to add up to X and all smaller number [1,X-1] could be obtained by adding some of numbers in series. to fulfill this I think the largest number have to be at max first integer number larger than X/2 for X=9-->largest number 5.

*Last edited by kappa_am (2014-04-07 07:50:11)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,588

I am getting

for 9 and 4. Any members you do not think belong there? If you agree with the two lists then we have a formula!

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**kappa_am****Member**- Registered: 2013-12-20
- Posts: 35

there is a condition all numbers up to x=9 could be obtained using the series member but: we cannot generate 4 using 6,1,1,1 or I cannot generate 1 using 2,2,3,2...

some members have to be obliterated.

according to above condition at least one room has to be 1.

Thank you for taking time to help me.

*Last edited by kappa_am (2014-04-08 03:10:35)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,588

I am not following you. How can you ever generate 1 using 4 integers all greater than 0?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**kappa_am****Member**- Registered: 2013-12-20
- Posts: 35

sorry, I write it vague.

let me clarify:

I have X and N number of rooms

here are the conditions:

summation of all N number have to be X

and all integer number less than X ([1 X-1]) should be obtained using at least one subset of those numbers.

I can obtain 1 of any series that have at least one 1 in that.

*Last edited by kappa_am (2014-04-08 03:55:46)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,588

So, why can you not generate 1 from a subset of 6,1,1,1?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**kappa_am****Member**- Registered: 2013-12-20
- Posts: 35

we can generate ,but we cannot generate 4 and 5 of this set 6,1,1,1

we cannot generate 1 of subsets that haven't any 1 like 2,2,3,2

*Last edited by kappa_am (2014-04-08 04:12:48)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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But there are some in my list that can certainly get 1,2,3,4,5,6,7,8.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**kappa_am****Member**- Registered: 2013-12-20
- Posts: 35

yes, the list is ok! just the sets that haven't any 1 and sets that have 6 have to be obliterate.

(6 1 1 1), (1 6 1 1), (1 1 6 1), (1 1 1 6), (2 3 2 2), (2 2 3 2), (2 2 2 3), (3 2 2 2) have to be obliterated.

I like a formula by having X and N, the number of sets can be obtained. it it obvious that at least one of rooms has to be 1. if we can find a rule for maximum number we may be able to find a formula.

the highest number for this example is 5. it has relationship with N and X. I am trying to develop a rule for maximum number. I think after that we could develop a rule for obtaining the number of sets.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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So you are saying there has to be a 1 in every set? And no 6?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**kappa_am****Member**- Registered: 2013-12-20
- Posts: 35

at least one 1 has to be in series because if it not be, I haven't any subset that its value is 1. this is general.

but 6 is for this specific example (x=9, N=4), because if 6 is in set all other member have to be 1 (minimum value). but in this situation I haven't any subset that add up to 4 and 5.

if x was 12 then 6 could be in series like (1,2,3,6).

I think if we could find a relationship for determining maximum value for each problem using N and X we could find the number of sets that satisfy the mentioned conditions.

Am I right?

*Last edited by kappa_am (2014-04-08 04:56:02)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,588

Am I right?

I do not know. Partition problems are tough especially when you have to leave certain numbers out and put others in.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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