Starting with a standard double-six set of dominos,
arrange the 28 dominos into the perimeter of a square, such that the pip values on each of the four
sides of the square sum to the same
value. At each corner of the square,
there is a 90
o
turn, so that each side is 7½ dominos long.
The usual rules apply; that is, where two dominos
meet, the ends must match (have
the same number of pips).
Place the
double value dominos lengthwise,
not crosswise. Present your answer
as four strings of 7½ dominos each,
starting at the upper left corner and
proceeding clockwise around the
square. For example, the top edge
might be: 0-3|3-3|3-5|5-5|5-0|0-2|2-4|4
(sum = 44), and the right edge might
be: 4-6|6-2|2-2|2-1|1-5|5-4|4-0|0 (sum =
44). There are multiple answers, and
any valid solution will be accepted.
There's more than one possible sum, the corner numbers must have the same parity and sum to a multiple of 4. (It's not possible for the 4 corners to be the same number.)
I played around with a set until I found this solution.
Corners 2,6,2,6. Sides sum to 46.
Top (l to r): 2|21|16|64|44|43|30|06|
Right (t to b): 6|62|25|54|41|11|13|32|
Bottom (l to r): |66|63|33|35|51|10|02|2
Left (t to b): |22|24|40|00|05|55|56|6
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Posted by Jer
on 2025-01-03 11:58:21 |
A solution
