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

EDIT: assuming all integers must be positive.

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Guest Message by DevFuse

Started by bonanova, Feb 20 2014 09:23 AM

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

Spoiler for I think this is the smallest

EDIT: assuming all integers must be positive.

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6 replies to this topic

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

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

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

*Vidi vici veni.*

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.**

Posted 22 February 2014 - 07:11 AM

Spoiler for in fact I can prove it

Posted 23 February 2014 - 12:17 AM

Spoiler for for largest common sum

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

Posted 23 February 2014 - 04:31 PM

Spoiler for found what Rainman is looking for

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.

*Vidi vici veni.*

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