If you look at all pairwise sums of the ten numbers given, five such totals occur exactly three times:
2475 = 1120 + 1355 = 1140 + 1335 = 1175 + 1300
2570 = 1120 + 1450 = 1140 + 1430 = 1270 + 1300
2605 = 1120 + 1485 = 1175 + 1430 = 1270 + 1335
2625 = 1140 + 1485 = 1175 + 1450 = 1270 + 1355
2785 = 1300 + 1485 = 1335 + 1450 = 1355 + 1430
Shortcut: If we assume that a < b < c < d < e, then we must have
a + b + c + d = 2475
a + b + c + e = 2570
a + b + d + e = 2605
a + c + d + e = 2625
b + c + d + e = 2785
which can now be solved to give
a = 480, b = 640, c = 660, d = 695, e = 790
as expected. |
