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I have four numbers 

x=50
y=30
z=20

sum of x+y+z=100

if i want to add a number n where

n=(x+y)/2

Edit:

   n=45
    such that x+y+z+n=100

now x+y+z=100.If any number (45 here) is added, then the sum should be always 100. i.e we need to subtract number(say m) from each x,y,z,n such that sum x+y+z+n should be equal to 100.

is it possible to do that.

Gerry Myerson
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Madhav Bhattarai
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  • in fact your equation is $3x+3y+2z=200$, now your question is that all natural's numbers such that satisfied this equation or not? –  Oct 11 '13 at 10:09
  • i want to maintain the weitage of number i.e if any number of variable is added to sum of x,y and z then the sum should always be 100. Also check the edit : n is having some value. – Madhav Bhattarai Oct 11 '13 at 10:15
  • note that if $n=(x+y)/2$ and $n=50$, then must be have $z=0$, i think your question must be refine. –  Oct 11 '13 at 10:21
  • I don't understand your question. You seem to have three numbers, not four, $x+y+z = 110 \ne 100$, and you are asking how to add $n=45$ to what? I assume you know how numbers are added to each other, so what is your question? – Vedran Šego Oct 11 '13 at 10:23
  • sorry for mistake just check the question i have edited, now x+y+z=100.If any number (45 here) is added, then the sum should be always 100. i.e we need to subtract number(say x) from each x,y,z,n such that sum x+y+z+n should be equal to 100. – Madhav Bhattarai Oct 11 '13 at 10:30
  • @user2749218 So, you want $(x-a)+(y-a)+(z-a)+n=100$, and your question is what is $a$? – Vedran Šego Oct 11 '13 at 10:48
  • yes exactly, you got it. – Madhav Bhattarai Oct 11 '13 at 10:50

1 Answers1

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Let $a = n/3$. Then

\begin{align*} (x-a) + (y-a) + (z-a) + n &= x - n/3 + y - n/3 + z - n/3 + n \\ &= x + y + z + n - n = x + y + z. \end{align*}

Vedran Šego
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