2
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    1
  8  9
 4 7 0
3 5 2 6

Arranged them so that the numbers in a way that the 3-sides added up alike-that is, to 16.

Can you rearrange the numbers so that the 3-sides shall sum to the smallest number possible? , Of course, the central barrel (which happens to be 7 above) is not a part of the count.

3-sides means 3 sides of the triangle...

1,8, 4, 3 is the first
1, 9, 0, 6 is second
and 3, 5, 2, 6 is the last...

@bonus question: - How many solutions will be there?

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4
  • $\begingroup$ what is a 3-side $\endgroup$ Sep 13, 2019 at 13:25
  • $\begingroup$ ? = 0 ${}{}{}{}$ $\endgroup$ Sep 13, 2019 at 13:26
  • $\begingroup$ @GeorgeMenoutis ok editing in more details.. wait $\endgroup$ Sep 13, 2019 at 13:27
  • $\begingroup$ @TheSimpliFire yes it is 0 $\endgroup$ Sep 13, 2019 at 13:31

1 Answer 1

4
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The smallest achievable sum is

13

Achieved as follows, one possibility

    0
  8  6
 4 9 5
1 7 3 2

Reasoning

Each of the numbers at the vertices of the triangle are used twice so we should make these as small as possible - namely, assign them to 0, 1 and 2. The number in the middle won't be used at all so we should make this as big as possible - namely 9. Then the sum of the three edges will be 2 $\times$ (0+1+2) + (3+4+5+6+7+8) = 39 so distributing equally, we should make each edge sum to 13.

Number of Solutions

This comes down to how many ways we can partition the set {3,4,5,6,7,8} into three sets of 2 so that the sums are 10, 11, and 12. There two ways of doing this: {(3,8), (4,6), (5,7)} and {(3,7), (4,8), (5,6)}. Within these groups we can swap each of the numbers around so each set corresponds to eight solutions which are geometrically distinct. Hence, there are 16 distinct solutions, up to combinations of rotations and reflections of the triangle.

If we count rotations and reflections as distinct solutions, this means there are 6 $\times$ 16 = 96 solutions.

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0

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