Extension: What is the smallest set (total weight) of five weights which, when taken in pairs, will produce 10 successive integral weights? Or can such a set be found? If not, find a set which produces integral weights, with as many successive integral weights as possible. (The above set has 7 successive integral weights, for example.)
Source: Henry Ernest Dudeny, in The Penguin Book of Curious and Interesting Puzzles, David Wells, 1992, #298. Used as Ken's POTD 10/25/94.
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
The total of these 10 weighings is equal to S = p + (p+1) +.....+(p+9) = 10*p + 45 with p equal to the lowest value. S is also equal to 4*sum(w_i). Since the pair wise sums of the w_i must be integers, it is easy to show that 2*sum(w_i) must also be an integer. Thus S is a multiple of 2; 10*p+45 cannot be a multiple of 2, so such a set of weights doesn't exist.