'Farmer Tomkins had five trusses of hay, which he told his man Hodge to weigh before delivering them to a customer. The stupid fellow weighed them two at a time in all possible ways, and informed his master that the weights in pounds were 110, 112, 113, 114, 115, 116, 117, 118, 120, 121. Now, how was Farmer Tomkins to find out from these figures how much every one of the five trusses weighed singly? The reader may at first think that he ought to be told "which pair is which pair" or something of that sort, but it is quite unnecessary. Can you give the five correct weights?'
Answer:
Let call w_1, w_2, w_3, w_4, w_5 the weights of the 5 trusses considered in
order of decreasing weight.
As there are 5 trusses, there are 10 possible ways ( 5x4/2) of weighting them
two at a time. As Hodge has got 10 different values, this means that the
weights of the 5 trusses are all different ==> w_1 > w_2 > w_3 > w_4 > w_5
As each truss is taken 4 times by Hodge, the total amount of the weights he
has measured represents 4 times the total of the weights of the 5 trusses.
Therefore sum(w_i) = (110 +112+....+121)/4 = 289
On the other hand, we have the following relations:
w_1 + w_2 = 121 and w_4 + w_5 = 110
Therefore w_3 = 58
As a result w_1 = 62 or 61 and w_5 = 53 or 54
It's easy to check that there is one solution defined by:
w_1 = 62, w_2 = 59, w_3 = 58, w_4 = 56, w_5 = 54
