bonanova Posted February 20, 2014 Report Share Posted February 20, 2014 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. Quote Link to comment Share on other sites More sharing options...

0 Rainman Posted February 21, 2014 Report Share Posted February 21, 2014 (edited) EDIT: assuming all integers must be positive. Edited February 21, 2014 by Rainman Quote Link to comment Share on other sites More sharing options...

0 Rainman Posted February 22, 2014 Report Share Posted February 22, 2014 Let the 16 numbers be n+1 through n+16, for some natural number n. Take the four rows of 5 numbers each, and add them together. You will get four numbers counted twice, and the rest counted once. The minimum total sum is (n+1)+(n+2)+(n+3)+...+(n+16)+(n+1)+(n+2)+(n+3)+(n+4) = 20n+146. Since all of these rows must have the same sum, the minimum sum for one of them is ceiling((20n+146)/4) = 5n+37. So the lowest possible sum per row is 37, for n=0, which is exactly what I found in my previous post. Quote Link to comment Share on other sites More sharing options...

0 superprismatic Posted February 22, 2014 Report Share Posted February 22, 2014 Here's a solution for the maximum common sum (41) using the integers from 1 to 16:....13..07.... ......09...... 04..11..14..12 ..06......03.. 08..16..15..02 ......10...... ....01..05.... I've found 8 distinct solutions (up to symmetries) for a common sum of 41. Proof that 41 is maximal: Let S be the common sum and let C be the sum of the four closest numbers to the center of the octagon. Then, since the numbers making C are used in three of the sums, whereas all the other numbers are in only two of the sums, 8*S - C = 32*n + 16*17 (n comes from the consecutive integers being n+1, n+2, n+3, ..., n+16). In this case, n=0. Now, we rearrange to solve for S to get S = 4*n + 34 + (C/8) which means that C must be divisible by 8. The largest C possible by adding four different numbers in the set {1,2,3,...,16} is 56. This makes S in our case to be 41. Quote Link to comment Share on other sites More sharing options...

0 Rainman Posted February 23, 2014 Report Share Posted February 23, 2014 OP doesn't require the numbers 1 through 16, just that the numbers are consecutive. Again, let the numbers be n+1 through n+16 for some natural number n. If we add together the four rows with four numbers each, we get four numbers counted twice, eight numbers counted once, and four numbers counted nonce. The maximum sum of this is (n+5)+(n+6)+(n+7)+...+(n+16)+(n+13)+(n+14)+(n+15)+(n+16) = 16n+184, which makes the maximum individual sum (16n+184)/4 = 4n+46. Considering that the minimum sum for a five number row is 5n+37, and the fact that the sums must be equal, we get 0 = (five number sum)-(four number sum) >/= (5n+37)-(4n+46) = n-9, which means n </= 9. However, we can't have equality because the numbers to be counted nonce in the four number rows are counted twice in the five number rows. For the five number rows to have a sum of 5n+37, or a total sum of 20n+148, the numbers counted twice must have a sum of 4n+12. But for the four number rows to have a sum of 4n+46, or a total sum of 16n+184, the same numbers must have a sum of 4n+10. This is a contradiction. So n<9. For n=8, I think it's possible to achieve a sum of 77 per row, but I have yet to find an example. Quote Link to comment Share on other sites More sharing options...

0 superprismatic Posted February 23, 2014 Report Share Posted February 23, 2014 Here are two such octagons for n=8. I believe there are no others up to symmetries.....18..20.... ......10...... 21..19..22..15 ..11......14.. 17..24..23..13 ......09...... ....16..12........20..21.... ......09...... 17..19..23..18 ..12......10.. 16..24..22..15 ......13...... ....14..11.... 1 Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted February 24, 2014 Author Report Share Posted February 24, 2014 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. Quote Link to comment Share on other sites More sharing options...

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## bonanova

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

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

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