While waiting at the Spaceport on Zargon for my long flight back to Earth from vacation, I did some souvenir shopping and found some strange looking dice used for weird board games known only to the Zargonians. These dice, each identical, had four sides instead of six, with each face arrayed into four triangular sections. While studying the dice, I noticed a few interesting things that might relate to how they're used in a game.
Each triangular section was numbered variously 1 through 9, with every digit except 1 and 9 appearing twice on each die (but on different faces). On each face, the center space was the sum of two of its corners, except on one face where it was the difference, and the centers comprised 4 consecutive numbers. The corners on one face summed to a square, on another to a prime number and the corners on the remaining two faces were each comprised of consecutive numbers.
I also noticed that only three odd numbers occurred around two of the vertices while only even numbers occurred around the other two, and in each case the three numbers were different. The product of one set of even numbers was a cube, the sum of the other three even numbers also equalled the sum of the digits in their product, and only one odd number was duplicated between the two odd vertices. Further, the sums of the three numbers at each vertex result in 4 more consecutive numbers.
What were the numbers on each face and how were they arranged?
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For convenience, present the solution as an unfolded die thusly. Your arrangement may vary from the solution (what you indicate as the center face, for example), but the contents and relationships of the faces won't change.
_
/*\
/ * \
/* *\
/-------\
/*\* */*\
/ * \ * / * \
/* *\*/* *\
Well, I'm afraid there is a little glitch when the problem states that "The corners on one face summed to a square, on another to a prime number and the corners on the remaining two faces were each comprised of consecutive numbers". In my early attempts I always read this as "one face sums to a square, another face sums to a prime and there are two other faces whose sums are neither square or primes but their numbers are consecutive (in each face)."
But I tried to solve it that way and it proved to be a dead end. That is, unless i repetead any mistake every time... So, my solution is similar to Daniel's:
/9\
/ 8 \
/2 6\
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/8\4 2/4\
/ 6 \ 5 / 7 \
/1 7\3/5 3\
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As you may see, there's a square (8+7+1), there's a prime (9+6+2), and there are two other faces with consecutive numbers in the corners. But wait! One (4+3+2) also results in a square!! That's the glitch I was talking about.
My solving goes through a little different way than Daniel's. Instead of listing all possible combinations for the dice faces, I went sorting out the numbers that could not be in a face, because they were already to be used up on the other faces, for instance, or because they could not sum up to a prime or to a square, or because these prime or square combinations didn't have two numbers that either added up or whose difference resulted in the centre number.
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Posted by vj
on 2008-12-05 12:18:59 |
i'm afraid there is a little glitch, but here it goes
