The Trusses of Hay

'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?'

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.

Solutions were received from many people: Denis Borris, Ravi Subramanian, Larry Baum, Nick Baxter, Radu Ionescu,, Joel Haywood, Philippe Fondanaiche, Israel Eisenberg, B Vijayavardhan.

  1. The weights for the first problem are 54,56,58,59,62. A representative solution (by Philippe Fondanaiche) is:
    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
  2. There is no set of five weights that can give 10 successive integers. From Nick Baxter:
    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.
  3. Weights can be found to give 9 successive integers. The lowest positive integral weights which do this are 1,2,3,5,8. The lowest positive weights are 1/2,3/2,5/2,9/2,15/2. The lowest weights which yield positive integers are 0,1,2,4,7.

Mail to Ken