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Fill in each blank white hexagon with a digit from 1 to 9. Each of the grey hexagons is the sum of the white hexagons around it, and all of the white hexagons that are around the same grey hexagon must be distinct.

The answer's already available, but I would like to see how people would logically solve this.

Source: USAMTS

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Building on shoover's answer, solving it step by step:

  • $30$ has 3 options, $[9, 8, \color{gray}{5}, 4, \color{gray}{3}, 1]$$[9, 7, \color{gray}{5}, 4, \color{gray}{3}, 2]$ and $[8, 7, 6, \color{gray}{5}, \color{gray}{3}, 1]$ (with gray indicating already known). As $4$ and $6$ already exist around $21$, the box shared by $21$ and $30$ must be either $1$ or $2$.

  • $26_{top}$ has 3 options, $[9, \color{gray}{6}, 5, 3, 2, 1]$, $[8, \color{gray}{6}, 5, 4, 2, 1]$ and $[7, \color{gray}{6}, 5, 4, 3, 1]$. Because $5$ already exists around $39$ the first option isn't applicable, and the two shared boxes with $39$ must be $4$ and $8/7$.

  • This means that the bottom of $39$ must be $9 + (8/7)$, depending on which one is around $26_{top}$. There is no solution for $26_{bottom}$ if it is $9 + 8$, so it must be $9 + 7$. This means that $8$ is around $26_{top}$, and $26_{bottom}$ is $[9, 7, 4, 3, \color{gray}{2}, 1]$.

                                                      

  • In all solutions for $30$ at this point, the only value it can share with $26_{bottom}$ beyond $9/7$ is $4$.

  • Since $32$ already has a $1$, we can place the remaining numbers around $26_{bottom}$.

  • Looking at the empty shared box between $32$ and $26_{top}$, the solution for $26_{top}$ dictates that this must be $5/2$. No solution exists for $32$ if its top-left box is $4$, so we can now place the $8/4$'s.

                                                      

  • There's no solution for $32$ if it is $9 + 2$, $7 + 5$ or $7 + 2$, so it must be $9 + 5$ with a solution of $[9, \color{gray}{8}, 6, 5, \color{gray}{3, 1}]$. This also places the $9/7$'s and the remaining around $32$.

  • The remaining boxes around $26_{top}$ are $2 + 1$. The only solution for $29$ is $[9, \color{gray}{6, 5, 4}, 3, 2]$, so their shared box must be $2$. This places the remaining around $26_{top}$.

  • The remaining around $29$ are $9 + 3$. As $21$ cannot have a $9$, their shared box must be $3$. This places the remaining around $29$.

                                                      

  • The remaining around $30$ are $9 + 2$. As $21$ cannot have $9$, this places the remaining blocks.

                                                      

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To get it started:

  • $21$ has to be $1+2+3+4+5+6$
  • $39$ has to be $9+8+7+6+5+4$
  • $21$ already has a $4$, so the overlap between $21$ and $39$ has to be $5$ and $6$, but $29$ already has a $5$, so it must get the $6$, while $30$ shares the $5$. (The hex shared by $21$, $30$, and $39$ is $5$, and the hex shared by $21$, $29$, and $39$ is $6$.)
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1
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This was really fun! After working it process of elimination after identifying max and min sums you can get from 6 digits 1-9, I was indeed able to solve.

I solved by saving image and notating like I would a Sudoku puzzle with possible choices. When in doubt, I would take combinations and choose absolute smallest or largest combinations between two hexagons and see if the rest of the cells were workable. This eliminated quite a few possibilities.

https://imgur.com/a/aIH1VLw

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  • $\begingroup$ My apologies if you were asking for a mathematical proof, btw! $\endgroup$ – Carley May 9 at 9:15

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