Friday, 5 June 2015

mathematics - Can you pass through the ancient tomb? Find the correct interpretation for the symbols


You have been crossing this hot and wet place for weeks now, but when you enter the cool and mouldy ruin, you know in seconds it has been worth it. Nobody has ever been here ( - well nobody in last millennia at least - ) and the ancient writings and symbols all over the place are comparable to none you've ever seen. Could it really have been build by aliens? The answer is quite possibly behind this huge, stone door in front of you. You approach cautiously, anxiously noticing the holes in the floor and ceiling which don't promise much good for unwelcomed guests. There is no handle on that door, but 5 oddly shaped intentions which are far too regular to be anything else than a hidden mechanism. enter image description here


And of course you recognize the shape immediately. After all, it was the amulet consisting of 5 elements shaped exactly like this, which send you on this endless quest in a first place. Eagerly, you pour out the pouch you've brought to bring out the five pieces which were once, ages ago, stringed together on a leather string. 5 indentations, 5 puzzle pieces of matching size - what could be more obvious?


enter image description here


But the one thing, which has been missing whenever you pondered over the meaning of the symbols on the pieces in the past is only now in front of you. Directly above the little panel, nearly aged away by the forces of nature over the years, are some instructions in the stone. Will they be enough to help you? Again you suspiciously eye the holes in the floor and the ceiling. There might only be once way to find out - but there might as well only be a single chance to get it right...


enter image description here


Labelling the 5 pieces from left to right as A, B, C, D, and E. In which order will you dare to put them into the 5 indentations (#1 to #5 from left to right.)?




I would like to thank Peter for sorting out an ambiguity of the original puzzle (and solving it as well). The puzzle as presented above does include this fix.




Answer




A D E B C



Let me begin by saying that next time you write a puzzle like this for people to submit answers via a computer keyboard, please please please use letters which we can type rather than weird symbols. I know it looks cooler, but it's going to make it hard to follow my explanation because there's no way I'm going to embed hundreds of images in this answer.




I start by assuming, based on the diagrams at the top of the key, that



the U symbol is addition, zigzag is division, spiral is multiplication, and the leaking container is subtraction.




Then each of the lines contains a trapezium or two superimposed trapezia, which I take as



balanced: equals symbol; unbalanced: an inequality symbol, to be discussed in more detail later.



The second line tells us that



the S-like symbol is the digit 4.



The fifth line tells us that




either the most significant digit in a two-digit number comes to the left and the symbol which I call L is a 1, or the most significant digit comes to the right and the symbol which I call p is a 1.



The eleventh line would then be information-free unless we assume the first of those alternatives, so I make that assumption.


The ninth line is rather suggestive:



It seems that p must divide into pS, which means that p must divide into S, but since S = 4 and we already have a 1 and a 4 that means that p = 2.



Now the penultimate line comes into play. I'm not aware of any human culture which uses number bases greater than 60, so searching up to base 60 for



cubes of the form n*n*n = 2*base + n we get the options 3 * 3 * 3 = 23 (base 12) or 5 * 5 * 5 = 25 (base 60). But with the first line LI / S = n and the sixth line D * D = nI we can look for two-digit squares nI such that 4 * n = base + I and this rules out base 60. Base 12 we get 10 / 4 = 3 and 6 * 6 = 30.




At this point we have a few digits and a few remaining lines to process.



I <=> 0
L <=> 1
p <=> 2
n <=> 3
S <=> 4
m <=> 5
D <=> 6

1. 10 / 4 = 3

2. 3 + 3 + 3 + 3 = 4 * 3
3. Q - 2 = 5
4. e * 6 (inequality) Q * b
5. b + 5 = 12
6. 6 * 6 = 30
7. 5 * 5 = 21
8. 2 - Q = w
9. Y + 6 = 24 / 2
10. d + Y = T
11. e + 6 (inequality) 14

12. 3 * 3 * 3 = 23
13. mirror-C (inequality) C



So those give us



Q = 7, b = 9, w = -5, Y = 8. The symmetry between m and w and their values being negative is very suggestive, and helps explain why amulet A doesn't appear at all but its mirror image does. It also fits nicely with 0 being the only horizontally symmetric digit. Equation 10 checks out: -9 + 8 = -1.



We still need to tackle the unbalanced trapezium.



Line 4 tells us that e (inequality) ten and a half and line 11 tells us that e (inequality) ten. If the inequality is < then e must duplicate a digit value for which we already have a symbol. So either the inequality is <= and e is the digit ten, or the inequality is > and e is the digit eleven. The latter fits with the reading of the unbalanced trapezium as a scale which tips down on the heavier side, and also with the superimposed balanced-and-unbalanced trapezia of the last line, which then means <=. So we have e = eleven and C is the only unassigned digit, ten. We resolve its sign with the aid of the last line: C is the positive, and the negative is mirror-C which is smaller than it.




So the amulets are



A: -11
B: 5
C: 6
D: -10
E: 1
and should be sorted in ascending order (the trapezium points right rather than left), giving solution A D E B C.




Note that this answer has been revised in accordance with the revision to the question, following some discussion with BmyGuest about sharpening up both of them. See the revision history for more details.


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