Puzzle 9: The Flames of Judgement

Puzzle Walkthrough

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The first vote revealed was for Zoey. So the parchment in the first position targeted Zoey.

Leo’s parchment was read before Bella’s, but not immediately before. This means Leo’s vote appears earlier in the sequence than Bella’s, with at least one vote between them.

Mina voted for the same person as Dante, though their parchments were not consecutive. So Mina and Dante share a target, and there is at least one other parchment between their positions.

The third vote revealed was for Bella. So the parchment in the third position targeted Bella.

Bella’s parchment was not the final one read, and she did not vote for Mina. So Bella is not in the fifth position, and when we place her, her target will not be Mina.

Zoey did not vote for Leo. When we place Zoey’s parchment, it cannot target Leo.

The fourth parchment was cast by the same person who received three votes total. This tells us that the voter in the fourth position is the contestant who ends up receiving three votes against them.

Dante’s parchment was read after Zoey’s but before Mina’s. So in the vote order, Zoey appears before Dante, and Dante appears before Mina.

The eliminated contestant was not Zoey. Therefore the contestant who gets the most votes cannot be Zoey.

The final parchment revealed voted for the same person as the third. Since the third vote was for Bella, the fifth vote was also for Bella.

The fourth parchment voted for someone whose name had already appeared in another vote. So the target of the fourth vote must be a name that appeared in positions one, two, or three.

Now connect the count logic. We already have two votes for Bella at positions three and five. The contestant in position four is the one who received three votes in total. That person cannot be Zoey, because the eliminated contestant was not Zoey. The only way to reach three votes cleanly with the information we have is for Bella to be the contestant who received three votes. That makes Bella the voter in position four. Because no one can vote for themselves, Bella’s fourth-position vote cannot target Bella. It must target a name that has already appeared. The first vote was for Zoey, so Bella’s vote at position four can target Zoey, satisfying the “already appeared” condition and Bella’s restriction of not voting for Mina.

We now have two votes for Zoey (first and fourth) and two votes for Bella (third and fifth). To bring Bella to three total, the second vote must also be for Bella.

Place the voters to satisfy the remaining ordering rules. Bella’s parchment is fourth by the earlier deduction. Leo’s parchment must be before Bella’s and not immediately before, so Leo must be first or second; it cannot be third. Since the third vote is for Bella and needs a voter, and since the second vote is also for Bella, we can fit the remaining identity and sequencing constraints as follows.

Make Leo the first voter who targeted Zoey, which places Leo before Bella and not immediately before. Zoey’s parchment must appear before Dante’s and Dante’s before Mina’s. Put Zoey second voting for Bella. Put Dante third voting for Bella. Put Bella fourth voting for Zoey. Put Mina fifth voting for Bella. Mina and Dante voted for the same person, Bella, and their parchments are not consecutive. Dante is after Zoey and before Mina. Zoey did not vote for Leo. Bella’s parchment is not the final one and did not target Mina. The fourth parchment targeted a name already seen earlier, Zoey.

Answers

Leo’s parchment was read 1st and he voted for Zoey.
Zoey’s parchment was read 2nd and she voted for Bella.
Dante’s parchment was read 3rd and he voted for Bella.
Bella’s parchment was read 4th and she voted for Zoey.
Mina’s parchment was read 5th and she voted for Bella.

Bella was eliminated.

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