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| Five numbers and Six sums (Posted on 2020-12-20) |
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A set of five numbers are added in all 10 possible pairs like in some of the Pumpkins puzzles.
When sorted the middle six sums are 47, 48, 50, 51, 52, and 54. The two largest sums and two smallest sums are not given, similar to Permuted Sums.
What are the possible sets of the five original numbers?
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Submitted by Brian Smith
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Rating: 5.0000 (1 votes)
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Solution:
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If there was a duplicate among the five original numbers then there would be a duplicate sum in
the middle six sums. But the given values are distinct so there are no duplicates in the
original numbers.
Let the numbers be A,B,C,D,E in ascending order. Then the six sum values given are A+D, B+C, B
+D, C+D, A+E, B+E in some order. There are some relations between the six expressions that hold
regardless of the specific values:
B+D>A+D, C+D>A+D, A+E>A+D, B+E>A+D, B+D>B+C, C+D>B+C, B+E>B+C,
C+D>B+D, B+E>B+D, B+E>A+E.
From these inequalities we can conclude:
A+D equals one of {47,48}
B+C equals one of {47,48,50}
A+E equals one of {48,50,51,52}
B+D equals one of {50,51}
B+E equals one of {52,54}
C+D equals one of {51,52,54}
Consider the sum A+B+D+E. This can be broken into two different groupings with the sums we have:
(A+D) + (B+E) and (A+E) + (B+D). The value A+B+D+E can only be a common value from the possible
sums of (A+D) + (B+E) and (A+E) + (B+D).
(A+D) + (B+E) = A+B+D+E | (A+E) + (B+D) = A+B+D+E
47 + 52 = 99 | 48 + 50 = 98
48 + 52 = 100 | 51 + 50 = 101
47 + 54 = 101 | 52 + 50 = 102
48 + 54 = 102 | 48 + 51 = 99
| 50 + 51 = 101
| 52 + 51 = 103
There are four possible matches that can be made. In each case the two unused values from
{47,48,50,51,52,54} can be assigned to B+C and C+D. Then our four cases can be summarized as:
Sum | Case 1 | Case 2 | Case 3 | Case 4
A+D | 47 | 47 | 47 | 48
B+C | 50 | 48 | 48 | 47
A+E | 48 | 50 | 51 | 52
B+D | 51 | 51 | 50 | 50
B+E | 52 | 54 | 54 | 54
C+D | 54 | 52 | 52 | 51
Each of these cases is a system of six equations in five variables. However one equaiton is
guaranteed to be redundant since these were generated guaranteeing (A+D) + (B+E) = (A+E) + (B+D).
The systems are not hard to solve (especially when using a computer to do row reduction on an
augmented 6x9 matrix). Then four solution sets are generated:
Case 1: 19.5, 23.5, 26.5, 27.5, 28.5
Case 2: 19.5, 23.5, 24.5, 27.5, 30.5
Case 3: 20, 23, 25, 27, 31
Case 4: 21, 23, 24, 27, 31 |
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