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?
 A: Hint:
To maximise $x$, we need to pay attention to the condition $x + e = 119$. The maximum possible is $x = 118$ as $e ≥ 1$, and with this value of $x$, it is possible to satisfy the other conditions (try it yourself!).
Then to minimise $x$, the largest three numbers should be consecutive, which results in $96 ≤ x$. When $x = 96, e = 23$, but then the remaining number is $320 - 23 - 283 = 14$, so $23$ is no longer the smallest number and $14 + 96 \ne 119$. Hence we need $283 + e = 320 - d$, and in the optimal case where $d = e+1$, what value of $x$ do we need?
A: $e\ge1,d>e\implies d\ge2$ 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?
A: The answer is 17. The maximum possible value of x is 118 and the least possible value is 101.
Explanation:
taking reference from the equations you use, I know:
$d + e = 37, d > e, d\neq e$
$e$ can range from 1 to 18. Now, using this information and the equation
$x + b + c = 283, x > b > c, x \neq b \neq c$
$x + e = 119$
Case 1: $e = 1$
$x = 118$, it satisfies the condition $x > b > c$ and there exists a solution where $c > d(=36)$
Case 2: $e = 18$
$x = 101$, it satisfies the condition $x > b > c$ and there exists a solution where $c > d(=19)$\
$\therefore$ the difference between the greatest value and the least value is 17 = (118 - 101)
