 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Pumpkins 6 (Posted on 2016-01-22) Much like the third Pumpkins puzzle six pumpkins, each having a different weight, are weighed two at a time in all 15 sets of two. This time exactly four pairs of duplicate values occurred. Those four values are 60, 100, 110, and 120 pounds.

How much did each individual pumpkin weigh?

 See The Solution Submitted by Brian Smith Rating: 4.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) re(5): Analytical solution (Bonus puzzle question) | Comment 9 of 11 | (In reply to re(4): Analytical solution (Bonus puzzle question) by Brian Smith)

I'm still working.

I confirm that there are no solutions for {100,110,116,120} where a+d = b+c and c+f = d+e.

I do have solutions to other problems where b+c = 100 and c+f = d+e = 220.

(38,48,52,58,62,68) gives {100,106,110,120}
(39,47,53,59,61,67) gives {100,106,114,120}

The complementary solutions are obtained by subtracting each pumpkin from 110 and their totals from 220:
(42,48,52,58,62,72) gives {100,110,114,120}
(43,49,51,57,63,61) gives {100,106,112,120}

None of these are Brian's bonus problem.  And, even if they were, I am not satisfied as I would not have demonstrated uniqueness.

I need a better analytical hammer.

 Posted by Steve Herman on 2016-01-30 09:36:23 Please log in:

 Search: Search body:
Forums (0)