Ok folks, so the three overlapping ellipses form seven curved regions. Your job is to place all seven tiles so that each region has exactly one tile and each ellipse adheres to the corresponding statement outside. Can you prove a unique solution?

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1 Answer 1




- Let us label the tiles as A-G first.
- $9$ must be present among $\{D,E,F,G\}$ otherwise even the largest chosen numbers wont add up to $20$.
- If $F=9$, then $\{B,C,D,E\}$ = $\{3,4,5,6\}$ and there cannot be a valid $A$ because $B+E+F > 15$. Similarly if $G=9$, then $\{A,F\}$ = $\{1,2\}$ and you cannot form $15$ using $A,B,E,F$ anyway.
- If $E = 9$, $\{A,B,F\}$ = $\{1,2,3\}$ and $\{C,D,G\}$ = $\{4,5,6\}$ but then $B+C+D+E > 18$ so it is not possible either. Thus, $D = 9$ must hold.

- We have three even-numbered tiles so the "adds to 18" and "adds to 20" sets must contain either zero or two even numbers. If zero even numbers then $\{B,C,D,E\}$ = $\{1,3,5,9\}$ and $A,F$ must be even which means $A+F+B+E$ is even which cannot be possible. So both the sets must contain exactly two even numbers.
- But we have three even tiles so $E$ must be even. If $E=6$, $\{B, C\}$ = $\{1, 2\}$ and we cannot find valid values for $F, G$. Similarly if $E=4$, $\{B,C\}$ = $\{2, 3\}$ and then $\{F, G\}$ = $\{1, 6\}$ which means $A=5$, but there is no valid choice for $B$ and $F$ satisfying the "adds to 15" set. So by elimination, $E = 2$.

- If $\{B, C\}$ = $\{3, 4\}$, there is no valid $F$ and $G$. So $\{B, C\}$ = $\{1, 6\}$ and $\{F, G\}$ = $\{4, 5\}$ implying $A = 3$.
- $B+F = 10$ which gives us $B = 6$, $F = 4$ and therefore $C = 1$ and $G = 5$.

  • 2
    $\begingroup$ Was just about to post this exact solution!!! Aargh! $\endgroup$ Nov 13, 2022 at 11:14

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