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 Pumpkins 5 (Posted on 2016-01-06)
Five pumpkins are weighed two at a time in all possible combinations. The results of the weighings gives nine different values, similar to the second Pumpkins puzzle.

But this time the repeated weight is omitted from the list. The eight distinct weights are 24, 28, 30, 34, 36, 38, 44, and 50 kilograms.

Determine the weights of the pumpkins and the omitted weight.

 See The Solution Submitted by Brian Smith Rating: 3.0000 (1 votes)

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 Solution | Comment 1 of 2
Label the weights such that A<B<C<D<E
The smallest sum tells us A+B=24
The next smallest tells us A+C=28
The largest sum tells us D+E=50
The next largest tells us C+E=44
The other sums are indeterminate at this point.

Consider possible weights of A,B,C:
11,13,17
10,14,18
9,15,19
etc.

Consider possible weights of C,D,E
18,24,26
17,23,27
16,22,28
etc.

There are only two possibilities for C (in bold) to check: {11,13,17,23,27} and {10,14,18,24,26}

The first case is the one that works.  It turns out B+E=C+D=40

The second case does not.  There is no sum to 30 and no repeats.

 Posted by Jer on 2016-01-06 11:14:20

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