# Hexagonal sum filling

Q1: Fill this hexagon with numbers 1-19 (no repetition) such that the sum of every vertical and diagonal row is the same. (This should be easy)

Q2: Assume that one of these pre-solved cells was not given. If you are still able to solve the problem with a unique solution, do you know which cell it is?

Bonus: How many solved cells can you delete and still be able to solve it? • Q2 seems to contradict the Bonus or maybe I don't understand Q2? Q2 suggests to me that there is one (unique) cell that can be removed and it still is solvable while the bonus says there are multiple
– Ivo
Sep 30, 2015 at 12:23

Solution to Q1:

       3
19    17
16     7    18
2     1
12     5    11
4     6
10     8     9
13    14
15


Solution to Q2:

The 1 is unnecesary. The 19, 12 and 7 cells come as usual. Then, between the 10 and 18 cells we have three numbers which must sum 10, and the only possible choice (since 2 and 3 are taken) is 1, 4 and 5. Below the 2 there must be a 4 (since else the number below would have to be either 12 or 16 (already taken). Then, the number 1 must be where it is (instead of 5) or else the number below the 18 would have to be 7 (already taken).

Bonus:

After removing the 1, the 17 is also unnecesary. If the number were a 19, no number could go in place of the actual 19. If the number were less than 17, the three numbers between 10 and 18 would have to sum less than 10 (which is impossible if 2 and 3 are already taken).

Moreover,

The 2 is also unnecesary, as it is possible to complete the outer ring without it with a bit of thinking: the 17, 19 and 12 come with a similar reasoning as before. Then, the 13 and 15 must be there (there is no other pair of available numbers which sum 28, and if the order were reversed, there would be no solutions for the two numbers below the 19). Then the 14, 9 and 11 come naturally. Once the outer ring is filled, the inner ring is also unique: between the 12 and 14 there must be a pair {5,7} or {4, 8}, and between the 13 and 11 there must be a pair {6,8}. Therefore, the bottom number of the inner ring must be 8, and the rest of the numbers are trivial to get.

Furthermore,

Removing the 16 cell also yields the same unique solution (I checked by trial and error).

TL;DR;

The puzzle has a unique solution with only the cells 3, 10, and 18. It is possible that this is not the only combination of 3 cells that yields a unique solution, and it may be even possible that the puzzle has a unique solution with 2 cells (this is the minimum required for symmetry reasons). However, checking all combinations for 2 given cells is almost impossible without brute-forcing the puzzle.

Q1.

       3
19    17
16     7    18
2     1
12     5    11
4     6
10     8     9
13    14
15


Q2. No, because...

Bonus: the magic hexagon is unique up to reflection and rotation, so any two cells other than the central one suffice to give a unique solution.

The sum in each row has to be 38.

We have all numbers from 1 to 19 which add up to 190 divided by 5 rows in each direction.

With this information we can start filling the hexagon easily until 7 numbers are left:

      03
19    17
16    07    18
02    01
12    ??    11
??    ??
10    ??    09
??    ??
??


leaving the numbers 4 5 6 8 13 14 15 which we have to fit by intelligent guessing:

      03
19    17
16    07    18
02    01
12    05    11
04    06
10    08    09
13    14
15


As for Q2 I'm not sure if i understand it correctly, but removing any of the numbers 3 17 18 would still give a problem with a unique solution, as 2 of them still give us the third.

• For Q2, he meant that if you remove one of those given numbers (assuming you have not seen/solved the puzzle before), you would still get a unique solution. What is that number? If you removed 3, 17, 18 and assuming you've never seen this puzzle before, how would you know the total of the diagonals. Sep 30, 2015 at 11:58
• @Takeshi As i said by the information that you can only use each of the numbers 1 to 19 exactly once. Sep 30, 2015 at 12:00