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Pumpkins 4 (Posted on 2014-01-31) Difficulty: 3 of 5
This is in continuation of Pumpkins 3.

Six pumpkins - each having a different weight - are weighed two at a time in all 15 sets of two. The weights are recorded as: 30, 33, 39, 45, 48, 51, 54, 57, 60, 63, 69, 75 and 78 pounds.
  • Precisely one of the values occurred exactly three times, but that value was only written down once.
Derive the weights of the pumpkins and which value occur thrice.

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
First solution (partial spoiler?) | Comment 1 of 3
Let the six pumpkins, smallest to largest, be a, b, c, d, e and f

Necessarily
30 = a + b
33 = a + c
78 = e + f
75 = d + f

Well, having learned from "Pumpkins 3", let's try the easy solution first.  One possible solution is:
39 = b + c
69 = e + d

This leads simply and directly to:
[a,b,c,d,e,f} = {12,18,21,33,36,42}
This proves to be a solution, as the pairwise sums are
30, 33, 39, 45, 48, 51, 54, 54, 54, 57, 60, 63, 69, 75, and 78

However, this is not necessarily the only solution.
It is possible that
39 = a + d
69 = c + f
This requires more work.
A full solution will follow.

  Posted by Steve Herman on 2014-01-31 10:09:58
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