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#1 2007-08-29 00:11:18

haruka-san
Member
Registered: 2006-11-08
Posts: 205

positive integral solution.

how to get the positive integral solution if
x= t +5
y= 5s-t
z= -3s-1

please give me the solution.

smile
thanks.
mwah! tongue


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#2 2007-08-29 01:52:40

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: positive integral solution.

Are you looking for something along the lines of 3x+3y+5z = 10?
I have no idea what a positive integral solution is.


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It wanted to be normal.

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#3 2007-08-29 05:16:36

Drew
Member
Registered: 2007-07-21
Posts: 2

Re: positive integral solution.

Yeah it would be useful if someone explains what a positive integral solution is.

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#4 2007-08-29 09:31:14

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: positive integral solution.

It means we want x, y and z to be all positive integers.

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#5 2007-08-29 10:34:38

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: positive integral solution.

Not sure of a general solution, but it should be obvious that s < -1/3 and t > -5.  Use these two restrictions on the middle equation and the answer should be easy.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#6 2007-08-29 22:36:56

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: positive integral solution.

It looks to me like the only solution that exists is s=-2/3, t=-4.
That's assuming that x, y and z can't equal 0.


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It wanted to be normal.

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#7 2007-09-05 07:27:51

Scratching head
Guest

Re: positive integral solution.

can anybody solve the problem ....find no of positive integral solutions of x1*x2*x3*x4=120

#8 2007-09-05 09:59:24

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

Re: positive integral solution.

Also t= -3 and s= -1/2 works for original question.
Then x,y,and z are all positive.

For Scratching head,  4 * 3 * 5 * 2 works.
2 * 6 * 5 * 2 works.
10 * 3 * 2 * 2 works.
15 * 2 * 2 * 2 is 120.
Can't think of others unless use a one (1) in the product.
Like 60 * 2 * 1 * 1, etc.


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#9 2007-09-05 22:41:00

scratching head
Guest

Re: positive integral solution.

but like can we factorise 120...and then from the factorised expression can we directly say??

#10 2007-09-06 00:07:20

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,711

Re: positive integral solution.

Well, we can find all factors here: All Factors of a Number

They are: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120

120 can be eliminated, as {1,120} are only 2 factors (and you specify 4)
likewise 60 {60,2,1} and 40 {40,3,1}
I think 30 also. {30,2,2,1} violates separate factors.
And maybe 24 can be eliminated also?

But 20 is in {20,3,2,1}

So we are left to play with 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#11 2007-09-06 01:07:01

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: positive integral solution.

John E. Franklin wrote:

Also t= -3 and s= -1/2 works for original question.
Then x,y,and z are all positive.

y and z aren't integers for that case though, and they are required to be.

About the second question, I'm not sure what the restrictions are. John's assumed that 1 isn't allowed, MathsIsFun is saying that all 4 numbers must be different. I believe the following is an exhaustive list, although it's possible that I've missed some:

{1,1,1,120}
{1,1,2,60}
{1,1,3,40}
{1,1,4,30}
{1,1,5,24}
{1,1,6,20}
{1,1,8,15}
{1,1,10,12}
{1,2,2,30}
{1,2,3,20} !
{1,2,4,15} !
{1,2,5,12} !
{1,2,6,10} !
{1,3,4,10} !
{1,3,5,8}   !
{1,4,5,6}   !
---
{2,2,2,15}
{2,2,3,10}
{2,2,5,6}
{2,3,4,5}   !

(If it's under the line then John likes it, if it's exciting then MathsIsFun like it. smile)


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