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my keyboard isn't working at the moment, it's not letting me input operations or equal signs. so i took a screenshot of the question :)

 Nov 16, 2018
 #1
avatar+106993 
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I am really not sure what the qustion is asking.

There are 3 unknowns,  x,y and z so how cha you have sloution pairs?

 

Here is my take.

\(x+2y+2z=n\\ x+2(y+z)=n\\ \text{Let } \color{red}Q=y+z\\ x+2Q=n\\ \)

 

Now i will solve for integer values of x+Q

I can see that  -1+2=1

so  1(-n)+2(n)=n

If k is an integer then

1(-n+2k)+2(n-k)=n

If x and Q are both positive integers then

 

\(-n+2k\ge1 \qquad and \qquad n-k\ge1\\ 2k\ge1+n \qquad and \qquad -k\ge1-n\\ k\ge \frac{1+n}{2} \qquad and \qquad k\le n-1\\ \frac{1+n}{2} \le k\qquad and \qquad k\le n-1\\ \frac{1+n}{2} \le k\le n-1\\ \text{Now I want 28 different possible integer values of k so }\\ n-1-\frac{1+n}{2}+1=28\\ n-\frac{1+n}{2}=28\\ \frac{2n-1-n}{2}=28\\ n-1=56\\ n=57 \)

 

But of course I have made up my own question because yours does not make sense to me.

If n=57 then there are  28 soltions for x an Q      [Not for x y and z]

 Nov 17, 2018

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