In a competitive sort of exam, the following question was aimed at 8th graders and above-
Sophie had written the numbers 1 to 22 in the 22 discs in the figure, but Adelaide, her big annoying sister, erased fourteen.
Find the positions of the numbers erased knowing that
- Each number written in the centre of a hexagon represents the sum of numbers placed at the vertices of this hexagon
- Two discs directly connected by a line never contain consecutive numbers.
On the answer sheet, write the values of the numbers $a, b, c, d$ and $e$.
On top of this, the question may have multiple answers (and the student is expected to report all of them) as well.
I don't have any idea about how to attack this problem. Of course, brute forcing is not a solution since in that case you have to check through $\sim 22^{14}\approx 6.2\times 10^{18}$ cases which is an impossibility even for a computer programme.
After doing some Google search recently, I found out something similar called Hexagonal Tortoise Problem and a paper about it. But, these couldn't take me anywhere.
One random observation is that if we indeed need to brute force the solution with some educated guesses, then maybe the disc connecting $4,7,12$ is the one to start with (since that's the one which is most restricted). But, even in that case, there are too many choices to deal with.
All ideas are welcome.
This question had an extremely elegant solution which was closed down because of the complaints of the question being a contest problem. The reality is that-
- I had already submitted my answers for the contest
- I posted the question only after the initial contest deadline
But unfortunately, I wasn't aware that the contest had extended their deadlines and hence I made this mistake.
I can't clearly remember the answer (except that it dealt with adding up the numbers in the boxes somehow). If someone can solve it, or someone here remembers it, or the actual original answerer sees it once more, please consider putting the answer here again.
Best Answer
The "elegant" solution you are missing is probably the following: Add all the numbers from the three hexagons at the corners. This gives on the one hand the sum of these three hexagons, on the other hand, this is equal to the sum of all numbers except 12, 16, b and c: $$69 + 56 + 66 = \sum_{k=1}^{22} k - (12 + 16 + b + c)$$ This yields $$b + c = 34.$$ Luckily, there is only one possibility to add two of the missing numbers to 34, namely b = 15 and c = 19 (b can't be 19 due to lying next to 20 in the grid). From there, it is easy to fill the remaining empty slots to obtain the solution that was presented in another answer.