The sum of five different positive integers is 320. The sum of greatest three integers in this set is 283. The sum of five different positive integers is $320$. The sum of the greatest three integers in this set is $283$. The sum of the greatest and least integers is $119$. If $x$ is the greatest integer in the set, what is the positive difference between the greatest possible value and the least possible value of $x$?
I obtained the equations
$$\begin{align}x+b+c+d+e &= 320\\
x+b+c &=283\\
x+e &= 119\\
d+e &= 37\\
b+c+d &= 201\end{align}$$
How to proceed after this?

$e\ge1,d>e\implies d\ge2$ and similarly $c\ge3, b\ge4$. Thus $x\le320-(1+2+3+4)=310$ from the first equation, $x\le283-(3+4)=276$ from the second and $x\le 119-1=118$ from the third. Since all three of these inequalities must be satisfied, the maximum value of $x$ is $118$. Can you verify by coming with with the values of the rest of the variables?
Suppose the minimum possible value of $x$ is $m$, then the highest possible sum of $x,b,c,d,e$ is $m+(m-1)+(m-2)+(m-3)+(m-4)=5m-10\ge320\implies m\ge66$. Similarly the second equation gives $m+(m-1)+(m-2)=3m-3\ge283\implies m\ge96$ and the third equations gives $m+(m-4)\ge119\implies m\ge62$. All three give $m\ge96$. Can you check that for $x=96$ we can find the values of other variables satisfying the equations? That is not the case: the problem occurs while choosing the value of $d,e$. We have $d=x-82,e=119-x$ so $d>e\implies 2x>201\implies x\ge101$. Does $101$ work?
