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
Puzzle Solution
