In the coinage of the People's Republic of Zinneravia, there are five types of coin, worth 10, 20, 30, 40 and 50 zinns.
You have an empty 4x4 grid like the one shown in the illustration below, and a good supply of each type of coin.
Given the following constraints…
- you can put one coin only in each square
- you can’t put any coin in a straight line (horizontally, vertically or diagonally) with another coin of the same value
… what is the maximum total value of the coins you can lay out in the grid?
Your answer should show a way of achieving the total in the following format:
[A, B, C, D] [E, F, G, H] [I, J, K, L] [M, N, O, P]
where each letter is replaced by
0 (no coin),
Show your workings. Only complete entry forms will be considered. The judge's decision is final.
This competition closed on Sunday 30th November 2014.
A lot of entries for this one! There was a roughly equal split into incorrect, partly correct, and fully correct.
The incorrect entries mostly involved missing the trick of using only three (not four) 40 zinn coins, which allows you to fill all the spaces with coins. The most obvious answer is 4x50, 4x40, 2x30, 2x20 and 2x10 zinn coins - leaving two empty spaces and totalling 480. Whereas the maximum is 4x50, 3x40, 3x30, 3x20 and 3x10 zinn coins totalling 500.
490 was another popular but sub-optimal solution.
The partly correct entries were in large part down to not following the instructions properly - examiners must despair! In particular (as you can see above) entrants were asked to state the total value, as well as an example solution in the format specified.
In the end there were a goodly number of right answers. (A couple of respondents devised smart computational solutions but most relied purely on logic.) Well done if you submitted a correct solution.
Here is the answer as given by our winner:
[10, 20, 50, 30] [50, 30, 40, 10] [40, 10, 20, 50] [20, 50, 30, 40] total value = 500
This solution has one set of four and four sets of three. To be improved there would need to be more sets of four, or some sets of five.
Sets of 5 are impossible from the starting empty grid.
A second set of four is possible, but then there cannot be any sets of three, so the optimal solution becomes two sets of four and four sets of two, which has a maximum value of 480.
And the winner is...
The winner of ecm's brainbuster no. 37 is Matthew Barr, who has a PhD in Nuclear Physics from the University of Birmingham. Matthew wins £200 of Amazon vouchers to spend on high-tech.
Congratulations again to Matthew, and watch out for ecm's next brainbuster!