Octo-Solution to “Evil Octo-Puzzler”
You may have picked up on the idea that the mummery attached to the last “Octo Puzzler” was merely a distraction. This is a logic problem, plain and simple and requires nothing more than a series of statements written out to solve it– no Octagon figure necessary, no bad Chuck Norris movie dialogue and no Ninjas. Sorry! Geometry makes everything look more complex than it really is!
What do we know: All numbers are whole (from the introduction). C is an even number (see clue 6). 1/5 E is even (see clue 3), so that means E is an even number. Thus E-C=20 (see clue 1, it’s either 20 or 21, and 20 is even). No number < 1 or > 35 (see the introduction), therefore E is a number in the range 22-34 and divisible by 5 (see above). Thus E=30, C=10, and G=5 (see clue 6). F=25 (see clue 7) and H-15 (see clue 4). D=20 (see clue 5), B=35 (see clue 2) and A-4 (see clue 8).
A couple of my favorite math problem prognosticators on Facebook stated this was too easy, that “once you deduced that A=4, everything falls into place”. I would have countered that that statement pretty much holds true for any of the sums.
Once you look upon the “clues” “introduction” and “hints” as logical filters, it is, in fact, rather easy. Tip of the hat to the lads (and ladies) of the West Seattle Gaming Group, for coming up with a unique solution:
Visit the MiniZinc math model site. Create a new MiniZinc model. Post in the following statement:
var 1..35: a;
var 1..35: b;
var 1..35: c;
var 1..35: d;
var 1..35: e;
var 1..35: f;
var 1..35: g;
var 1..35: h;
constraint e-c=20 \/ e-c=21;
constraint e=10 \/ e=20 \/ e=30;
constraint a+b+c+d+e+f+g+h=144;solve satisfy;output [show(a), “\n”,
Surprise! It solves the problem, based upon the filter set fed into it. Hope you enjoyed that, and Now it looks like I’ll have to shop around another one for January. You know, that will actually run in January.