With five numbers there are 5 ways to group 4 of them together. If you take the averages of each group of 4 numbers and get...

3428.5

3425.75

3398

3120.25

342.5

What are the 5 numbers?

Let the five numbers be A, B, C, D and E, with A<B<C<D<E.

Clearly, A+B+C+D must then be the lowest group sum followed by A+B+C+E, while B+C+D+E must be the highest group sum, preceded by A+C+D+E

Now, by the problem, we must have:

(A+B+C+D)/4 = 342.5 ....(i)

(A+B+C+E)/4 = 3120.25....ii)

(A+B+D+E)/4 = 3398......(iii)

(A+C+D+E)/4 = 3425.75....(iv)

(B+C+D+E)/4 = 3428.5.....(v)

Adding (i) thru (v), we obtain:

A+B+C+D+E = 13715…..(vi)

Accordingly, multiplying (i) thru (v) by 4 in turn and separately

subtracting these from (vi), we obtain:

(E, D, C, B, A) = (12345, 1234, 123, 12, 1)

Thus, the required numbers (in ascending order) are 1, 12, 123, 1234 and 12345

*Edited on ***August 25, 2007, 11:57 pm**