(Author Note) A while ago I made some puzzles for a now defunct puzzle hunt project, here's one of them. The answer to this (and most other puzzles of said hunt) is a single word. The puzzle itself is fully contained within the image above. Author notes don't matter.
1 Answer
The symbols are
We can convert to
the atomic number of each element
to transform the statements into
1. 6 (carbon) + 3⦶ = 50 (tin)
2. 82 (lead) + 🜔 = 83 (bismuth)
3. 50 (tin) + ⦶ = 82 (lead)
4. 51 (antimony) + ⦶ - 4🜔 = 79 (gold)
5. 47 (silver) - ⦶ = 29 (copper)
6. 77 (iridium) + 2🜔 = 79 (gold)
7. 79 (gold) - ⦶ = 47 (silver)
We can think of each of these statements as performing alchemy as follows:
Starting with one element, apply some transformations to obtain a different element.
Based on statements 2 and 6, we can infer that
Applying a multiple of 🜔 changes the element's atomic number by that multiple.
Turning our attention to statements 3, 4, and 7, it appears that
Applying a multiple of ⦶ changes the element's atomic number by that multiple of 32.
Finally, from statements 1 and 5, we learn that
Applying ⦶ still results in a large change, but with a smaller magnitude than 32.
These rules could be visualized if
we placed the elements in unequal rows based on atomic number and viewed 🜔 and ⦶ as horizontal and vertical movement, respectively.
Of course, there is a well known way to do this:
The periodic table.
This gives us the proper way to interpret the 7 statements:
1. Carbon (6 C) shifted down by 3 yields tin (50 Sn).
2. Lead (82 Pb) shifted right by 1 yields bismuth (83 Bi).
3. Tin (50 Sn) shifted down by 1 yields lead (82 Pb).
4. Antimony (51 Sb) shifted down by 1 and left by 4 yields gold (79 Au).
5. Silver (47 Ag) shifted up by 1 yields copper (29 Cu).
6. Iridium (77 Ir) shifted right by 2 yields gold (79 Au).
7. Gold (79 Au) shifted up by 1 yields silver (47 Ag).
Turning our attention to the unanswered statements at the end of the puzzle, we can convert and transform them as follows:
1. Bismuth (83 Bi) shifted right by 1 yields polonium (84 Po).
2. Silver (47 Ag) shifted left by 5 and down by 1 yields tungsten (74 W).
3. Samarium (62 Sm) shifted right by 6 yields erbium (68 Er).
Finally, we can
combine the resulting chemical symbols (Po, W, Er)
to obtain the answer:
POWER.
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$\begingroup$ Satisfying enough to be accepted, although you are missing a part to the puzzle with the salt and nitre. rot13(Znlor guvax bs gurz abg nf ahzoref, ohg irpgbef) $\endgroup$ Mar 14, 2022 at 8:09
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$\begingroup$ @Auribouros Thanks, I'll try to work it out. $\endgroup$– noedneMar 14, 2022 at 14:46
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$\begingroup$ As another hint (Gur irpgbef ner zrnag gb or hfrq fbzrjurer jurer gurer ner n ybg bs ryrzragf) $\endgroup$ Mar 30, 2022 at 11:32