# Problem of the Week Problem C and Solution Order Up!

## Problem

The letters $$w$$, $$x$$, $$y$$, and $$z$$ each represent a different positive integer greater than $$3$$. If we know that $\dfrac{1}{w-3}=\dfrac{1}{x+1}=\dfrac{1}{y+2}=\dfrac{1}{z-2}$ then write $$w$$, $$x$$, $$y$$, and $$z$$ in order from the letter that represents the smallest integer to the letter that represents the largest integer.

## Solution

Solution 1:

Since the fractions are all equal and they all have a numerator of $$1$$, that means that their denominators must all be equal. So $$w-3=x+1=y+2=z-2$$.

Now let’s suppose that $$w=10$$. Then $$w-3=10-3=7$$.

So $$7=x+1=y+2=z-2$$. We can make the following conclusions.

• Since $$7=x+1$$, that means $$x=7-1=6$$.

• Since $$7=y+2$$, that means $$y=7-2=5$$.

• Since $$7=z-2$$, that means $$z=7+2=9$$.

So when $$w=10$$, we have $$x=6$$, $$y=5$$, and $$z=9$$. We can see that $$x$$ is four less than $$w$$, $$y$$ is five less than $$w$$, and $$z$$ is one less than $$w$$. So when we write these in order from smallest to largest, we get $$y$$, $$x$$, $$z$$, $$w$$.

Solution 2:

As with Solution 1, we notice that since the fractions are all equal and they all have a numerator of $$1$$, that means that their denominators must all be equal.

So $$w-3=x+1=y+2=z-2$$. Let’s add $$3$$ to each expression.

$\begin{array}{c c c c c c c} w-3 & = & x+1 & = & y+2 & = & z-2\\ \textcolor{red}{\downarrow {\scriptstyle +3}} & & \textcolor{red}{\downarrow {\scriptstyle +3}} & & \textcolor{red}{\downarrow {\scriptstyle +3}} & & \textcolor{red}{\downarrow {\scriptstyle +3}} \\ w & = & x+4 & = & y+5 & = & z+1 \end{array}$

From this we can make the following conclusions.

• Since $$w=z+1$$, that means $$w$$ is $$1$$ more than $$z$$, so $$w>z$$.

• Since $$z+1=x+4$$, that means $$z$$ is $$3$$ more than $$x$$, so $$z>x$$.

• Since $$x+4=y+5$$, that means $$x$$ is $$1$$ more than $$y$$, so $$x>y$$.

So when we write these in order from smallest to largest, we get $$y$$, $$x$$, $$z$$, $$w$$.