The total is \(5\times8=40\). Let the smallest number be \(a\), so the largest is \(a+10\). Neither extreme can be 9: a smallest value of 9 makes the mean too large, and a largest value of 9 would make the smallest value negative.

There must be at least two 9s. With 9 in positions 2 and 3, write \(a,9,9,d,a+10\). The total condition gives \(2a+d=12\). Since \(d\geq9\) and \(a\geq1\), the only option is \(a=1,d=10\), giving \(1,9,9,10,11\).

With 9 in positions 3 and 4, write \(a,b,9,9,a+10\). Now \(2a+b=12\), with \(a\leq b\leq9\). The possibilities are \((a,b)=(2,8),(3,6),(4,4)\).

The last pair gives \(4,4,9,9,14\), where 4 ties with 9 as a mode, so reject it. The other two pairs give the stated sets.

Three 9s in the middle would give \(a+27+(a+10)=40\), or \(2a=3\), impossible for whole numbers. Each listed set has total 40, middle value 9, range 10 and unique mode 9.

Answer

The three possible ordered sets are:

\(1,9,9,10,11\); \(2,8,9,9,12\); \(3,6,9,9,13\).