Let $$a < b < c < d < e$$ be real numbers. Among 10 sums of the pairs of these numbers, the smaller three are 32, 36, 37, while the larger two are 48 and 51. Find all possible values of $$e$$.

The two smallest sums must be $$a + b = 32$$ and $$a + c = 36$$, the third smallest is $$a + d$$ or $$b + c$$, and the two largest are $$d + e = 51$$ and $$c + e = 48$$. Then $$c = 48 - e$$, $$d = 51 - e$$, $$b = c - 4 = 44 - e$$ and $$a = 32 - b = e - 12$$. If $$a + d = 37$$, then $$(e - 12) + (51 - e) = 39 \ne 37$$, impossible; so $$b + c = 37$$, giving $$92 - 2e = 37$$ and $$e = 27.5$$. This gives $$a = 15.5$$, $$b = 16.5$$, $$c = 20.5$$, $$d = 23.5$$, whose pair sums are 32, 36, 37, 39, 40, 43, 44, 44, 48, 51, as required. So the only possible value is $$e = 27.5$$.