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Ceremonial Stones (Posted on 2006-11-14) Difficulty: 5 of 5
A certain tribe had an interesting kind of annual festival, in which every male member of the tribe (regardless of age) had to contribute a levy of grain into the tribal store. Their unit of weight was roughly the same as our pound avoirdupois, and each tribesman had to contribute one pound of grain for every year of his age.

The contributions were weighed on the tribe's ceremonial scales, using a set of seven ceremonial stones. Each of these weighed an integral number of pounds, and it was an essential part of the ritual that not more than three of them should be used for each weighing, though they need not all be in the same pan.

If ever a tribesman lived to such an age that his contribution could no longer be weighed by using three or fewer stones, the levy of grain would terminate forever. And in the previous year, one old man had died only a few months short of attaining this critical age, greatly to the relief of the headman of the tribe.

It has been determined that the stones can measure the maximum age. What is this age and what were the weights of the seven ceremonial stones?

See The Solution Submitted by Bractals    
Rating: 3.0000 (7 votes)

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Hints/Tips A computer exploration -- solution??? | Comment 9 of 26 |

For speed I chose VB in preference to QB. (In particular VB 5.0).

Dim h(7), which, bestH(7), highTot
Sub try(lbs)
 h(which) = lbs
 If which = 3 Then
  Print h(1), h(2), h(3)
  Print highTot
  For i = 1 To 7
    Print bestH(i);
 End If
 If which = 7 Then
  ReDim had(200)
  For i = 1 To 7
   hi = h(i)
   had(hi) = 1
   For j = i + 1 To 7
    hj = h(j)
    hij = hi + hj
    himj = Abs(hi - hj)
    If hij <= 200 Then had(hij) = 1
    had(himj) = 1
    For k = j + 1 To 7
     hk = h(k)
     hijk = hij + hk
     hijmk = Abs(hij - hk)
     himjk = himj + hk
     himjmk = Abs(himj - hk)
     If hijk <= 200 Then had(hijk) = 1
     If hijmk <= 200 Then had(hijmk) = 1
     If himjk <= 200 Then had(himjk) = 1
     If himjmk <= 200 Then had(himjmk) = 1
  For i = 1 To 200
   If had(i) = 0 Then Exit For
  tot = i - 1
  If tot > highTot Then
   highTot = tot
   Print h(1), h(2), h(3)
   Print highTot
   For i = 1 To 7
    bestH(i) = h(i)
    Print h(i);
  End If
  which = which + 1
  For i = Int(lbs * 1.5) To lbs * 3.4
   If i <= 200 Then
    try i
   End If
  which = which - 1
 End If
End Sub

Private Sub cmdStart_Click()
  which = 1
For i = 1 To 4
  try i
End Sub

Early exploration with a QB program with wider ranges of possibilities indicated that good results could come with each successive weight being between 1.5 and 3.4 times the weight of the previous one, and that was incorporated into this search, as was a limit on the first, to be 1 through 4.

At the end, the set 1, 3, 9, 14, 40, 64, 97 allows the weighing to continue through 114 as the maximum age.

  Posted by Charlie on 2006-11-14 23:42:00
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