Hi all,
So I was on the road from the Bronx to Baltimore, struggling to stay awake, when I remembered the math problem that Dr. Yuen gave to us last night. Well, the first thing I realized was that she gave us the wrong problem.
So the problem is this: you have a 3x3 grid, where all vertical, horizontal, and diagonal 3-member lines must add up to 30. Dr Yuen, incorrectly, said that we needed to use all numbers 7->15 inclusive. This is clearly wrong because there is only one combination of numbers that will add up to 30 using 15: specifically, 7+8+15. Since you can't reuse numbers, and every single member of the 3x3 grid will be used in at least 2 lines, this is impossible.
I'm assuming that she meant we used the numbers 6 to 14 inclusive. This problem makes more sense. First off, we need to figure out what the question really is asking. If we think about it, there are 9 numbers, and 8 equations that must add up to 30. This could probably be solved computationally very simply, but that ruins the fun. There are too many degrees of freedom for this to be solved by number crunching.
However, we can realize that there are specific boundary conditions based on the numbers we have. Assuming 14 is the highest number, there are only 2 number combinations that will add up to 30: 14+7+9 and 14+10+6. Likewise, for 6: 6+11+13 and 6+10+14. Yay. Now if you think about it real quick, there are only 4 places on the grid that used in only two equations. If you realize that, the numbers fall into place:
7, 14, 9
12 ,10, 8
11,6,13
(When I realized this, I got so excited and almost got my 2nd speeding ticket in 3 wks. D'oh.) Dr. Yuen says that there is another combination that works. Hm. Maybe I can figure it out on the ride home. I don't have Dr. Yuen's email address though.
Sincerely,
Anonymous.
