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# Magic octagon

Best Answer Rainman, 21 February 2014 - 01:45 AM

Spoiler for I think this is the smallest

EDIT: assuming all integers must be positive.

Go to the full post

6 replies to this topic

### #1 bonanova

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Posted 20 February 2014 - 09:23 AM

By connecting the eight vertices of an octagon (with horizontal, vertical

and 45-degree lines, ignoring the perimeter) we define eight rows having

either four or five points of intersection.

Place sixteen consecutive (positive) integers on these points such that

1. The sums along all eight rows are the same.
2. The common sum is the smallest possible.

A harder problem is to create the largest possible common sum.

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Vidi vici veni.

### #2 Rainman

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Posted 21 February 2014 - 01:45 AM   Best Answer

Spoiler for I think this is the smallest

EDIT: assuming all integers must be positive.

Edited by Rainman, 21 February 2014 - 01:46 AM.

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### #3 Rainman

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Posted 22 February 2014 - 07:11 AM

Spoiler for in fact I can prove it

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### #4 superprismatic

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Posted 23 February 2014 - 12:17 AM

Spoiler for for largest common sum

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### #5 Rainman

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Posted 23 February 2014 - 01:15 PM

OP doesn't require the numbers 1 through 16, just that the numbers are consecutive.

Spoiler for upper bound for sum

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### #6 superprismatic

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Posted 23 February 2014 - 04:31 PM

Spoiler for found what Rainman is looking for

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### #7 bonanova

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Posted 24 February 2014 - 07:22 AM

I can mark only one post, so it's Rainman's, for getting the lowest sum.

Rainman and SP finished off the rest of the analysis correctly.

Nice work.

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Vidi vici veni.

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