Alpha Centipede
Member
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Alright, so I managed to get the solution without using a cipher, and using all the clues we got so far. This is the process I used: spoiler:
Starting from VOJWYJSUJGTGPNXWAFNKWNTTXTJTFZGGY, Clue #1 says to sort in a dead end number of rows. Dead end refers to the rare waves and the rare wave with the Leftovers had "THREE" scratched into the wall. So I sorted the long string of letters into three rows in the most natural way (at least to me), which is to take the first letter, in this case V, and then take every third letter from there. Doing the same thing with O and J, the second and third letters respectively, gives us the following three strings of length 11: VWSGPWNNXTG OYUTNAKTTFG JJJGXFWTJZY Putting them together as in clue #2 to obtain VWSGPWNNXTG OYUTNAKTTFG JJJGXFWTJZY and then reversing the resulting string as per clue #3 gives us: YZJTWFXGJJJGFTTKANTUYOGTXNNWPGSWV Now, clue #5 translated to alphabets gives us the string "SUBTRACT THE KEY". This also gives us a clue on how to go about dealing with our string of alphabets. Converting our letters from the 33-character alphabet string into numbers, we get 25, 26, 10, 20, 23, 6, 24, 7, 10, 10, 10, 7, 6, 20, 20, 11, 1, 14, 20, 21, 25, 15, 7, 20, 24, 14, 14, 23, 16, 7, 19, 23, 22 We also have clue #4. The answer is the key. The key. The key. The key. Repeated occurrences of the answer. And what is the answer? Well, Key already gave it to us before we went in. The answer is ZXJyb3I (or translated from base 64, "ERROR") We convert ERROR to the numbers 5, 18, 18, 15, 18. Now we use clue #5. Clue #5 tells us to subtract the key. And clue #4 tells us to do it repeatedly. So we subtract 5, 18, 18, 15, 18, 5, 18, 18, 15, 18, 5, 18, 18, 15, 18, 5, 18, 18, 15, 18, 5, 18, 18, 15, 18, 5, 18, 18, 15, 18, 5, 18, 18 from 25, 26, 10, 20, 23, 6, 24, 7, 10, 10, 10, 7, 6, 20, 20, 11, 1, 14, 20, 21, 25, 15, 7, 20, 24, 14, 14, 23, 16, 7, 19, 23, 22 to obtain 20, 8, -8, 5, 5, 1, 6, -11, -5, -8, 5, -11, -12, 5, 2, 6, -17, -4, 5, 3, 20, -3, -11, 5, 6, 9, -4, 5, 1, -11, 14, 5, 4 We have negatives here, so we wrap them around modulo 26 to obtain 20, 8, 18, 5, 5, 1, 6, 15, 21, 18, 5, 15, 14, 5, 2, 6, 9, 22, 5, 3, 20, 23, 15, 5, 6, 9, 22, 5, 1, 15, 14, 5, 4 which when converted back to alphabets, finally gives us THREEAFOUREONEBFIVECTWOEFIVEAONED Welp, that was a fun puzzle. I really loved it. Thanks, Verly!
< Message edited by Alpha Centipede -- 4/5/2020 15:33:49 >
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