How to find value of $x+y+z+u+v+w$

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

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

• Could you explain where do the numbers 1949 and 2004 come from? I am guessing that the former is the year of Chinease communist takeover, but 2004 is unclear to me. Oct 27, 2013 at 10:20
• $x$ is present in the LHS but not in the RHS. Don’t know if that remark is useful ... Oct 28, 2013 at 8:47
• It seems that $x=1,y=36,z=2324,u=1,v=5150,w=72130$ yields a solution, giving $x+y+z+u+v+w=79642$. Oct 28, 2013 at 10:51
• $x=1$. Let each side of equation = $2004*1949k$ where $k$ is positive integer. Multiplying the parentheses on right-hand-side with $x$ and subtracting from the parentheses on left-hand-side gives $xyzvw(u-1)+xyvw+zuvw+zu+zw+vw+1=(2004-1949x)k$. The left-hand-side is positive so right-hand-side must be positive so $x=1$. Oct 28, 2013 at 22:02
• You're right, I was incorrect. My bad. Oct 31, 2013 at 21:39

This is not a solution but a little progress towards the solution pointed out in Ewan Delanoy's comment to show that $x=u=1$ and $y \ge 36$.

Let each side of original equation be $2004 \times 1949 \times k$. Multiplying the parentheses on right-hand-side with $x$ and subtracting from the parentheses on left-hand-side gives $xyzvw(u−1)+xyvw+zuvw+zu+zw+vw+1=(2004−1949x)k$ . The left-hand-side is positive so right-hand-side must be positive so $x=1$.

If $x=1$,

$yzuvw+yzu+yzw+yvw+uvw+zuvw+y+u+w+zu+zw+vw+1=2004k$,

$yzvw+yzu+yzw+uvw+y+u+w=1949k$.

Since $z\ge 1$, first eqn yields $y(zuvw+zu+zw+vw)+2(uvw+u+w)+y+vw+1 \le 2004k$ and dividing by second eqn yields $\frac{y(zuvw+zu+zw+vw)+2(uvw+u+w)+y+vw+1}{y(zvw+zu+zw+1)+uvw+u+w} \le \frac{2004}{1949}$ which implies $\frac{zuvw+zu+zw+vw}{zvw+zu+zw+1} < \frac{2004}{1949}$ or $1949uzvw+1949vw<2004zvw+55(zu+zw)+2004$.

Substituting $zu \le uzvw$, $zw \le zvw$, and $1 \le vw$, $1949uzvw+1949vw<2059zvw+55uzvw+2004vw$ or $1894uz<2059z+55$ which precludes $u \ge 2$.

With $x=u=1$, the two equations reduce to

$yzvw+yz+yzw+yvw+vw+zvw+y+w+z+zw+vw+2=2004k$,

$yzvw+yz+yzw+vw+y+w+1=1949k$

and the difference of the two equations is

$yvw+zvw+z+zw+vw+1=55k$.

The ratio of the last two equations is

$\frac{(yz+1)(vw+w+1)+y}{(z+1)(vw+w+1)+yvw-w} = \frac{1949}{55}$.

But $y \le yvw - w$ unless $v=w=1$. Even if $v=w=1$, $y/(yvw - w) \le 1959/55$ so

$\frac{yz+1}{z+1} \ge \frac{1949}{55}$ so $y \ge 36$.

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 $w = 3$.