 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Pumpkins 4 (Posted on 2014-01-31) 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.) Solution, with no guesswork! Comment 3 of 3 | Call the pumpkins A-F, in ascending order of weight.

A+B must be the smallest weight and A+C must be the second smallest.  Similarly, E+F must be the largest weight and D+F must be the second largest.

The triple weight must be A+F = B+E = C+D.  The six pairs A+B, A+C, A+D, A+E, B+C, and B+D must all be less than this weight.  Similarly, the six pairs B+F, C+F, D+F, E+F, C+E, and D+E must all be more than this weight.  With only thirteen weights listed the triple weight must be the median weight.

A set of redundant equations can be formed:
A+B = 30
A+C = 33
D+F = 75
E+F = 78
4A+3B+2C+2D+E = 30+33+39+45+48+51
B+2C+2D+3E+4F = 57+60+63+69+75+78
A+F = 54
B+E = 54
C+D = 54

Solving this system yields that the pumpkins individually weigh 12, 18, 21, 33, 36, and 42 pounds.

Note that the first six equations alone can determine the pumpkins' individual weights, so the problem could have stated that the triple weight was completely left off the list and the problem would still be solvable!

 Posted by Brian Smith on 2015-12-24 11:48:41 Please log in:

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