TITLE: How to find value of $x+y+z+u+v+w$
QUESTION [8 upvotes]: let $x,y,z,u,v,w$ be positive integer numbers,and such
$$1949(xyzuvw+xyzu+xyzw+xyvw+xuvw+zuvw+xy+xu+xw+zu+zw+vw+1)=2004(yzvw+yzu+yzw+uvw+y+u+w)$$
Find this value of
$$x+y+z+u+v+w=?$$
My try:
maybe use

$$(x+1)(y+1)(z+1)(u+1)(v+1)(w+1)=(xyz+xy+yz+xz+x+y+z+1)(uvw+uv+uw+vw+u+v+w+1)$$

REPLY [2 votes]: It seems the following. 
I found the original problem (second picture, Problem 15).
Unfortunately, the problem expression is long and complex and there are misprints in your version. The right formulation should be:   
$$1949(xyzuvw+xyzu+xyzw+xyvw+xuvw+zuvw+xy+xu+xw+zu+zw+vw+1)= 2004(yzUvw+yzu+yzw+YVW+uvw+y+u+w).$$
The problem was solved by thepiano (here or here) and weiye (here).
Here is weiye’s solution. Build the following continued fraction representation: 

The uniqueness of the continued fraction representation implies that the values of variables are determined by the continued fraction representation of $2004/1949$. Therefore $x = 1$, $y = 35$, $z = 2$, $u = 3$, $v = 2$, and $ w = 3$.