You are not logged in.
Pages: 1
how to get the positive integral solution if
x= t +5
y= 5s-t
z= -3s-1
please give me the solution.
thanks.
mwah!
Dont compare your life to others. You have no idea what their journey is all about. ^_^
Offline
Are you looking for something along the lines of 3x+3y+5z = 10?
I have no idea what a positive integral solution is.
Why did the vector cross the road?
It wanted to be normal.
Offline
Yeah it would be useful if someone explains what a positive integral solution is.
Offline
It means we want x, y and z to be all positive integers.
Offline
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..."
Offline
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.
Why did the vector cross the road?
It wanted to be normal.
Offline
can anybody solve the problem ....find no of positive integral solutions of x1*x2*x3*x4=120
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.
igloo myrtilles fourmis
Offline
but like can we factorise 120...and then from the factorised expression can we directly say??
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
Offline
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. )
Why did the vector cross the road?
It wanted to be normal.
Offline
Pages: 1