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For this task you will be presented four lists consisting of eight items. All items consist of two parts. To make things slightly easier for you, the length of the shortest part is given.

In each list the items share a common theme; your task is to identify the odd one out. Once you have completed this task, you may notice that all the odd ones out again share a common theme. Identify the odd one out once more and notice how that fits in with the grand theme of this puzzle.

I trust I have informed you sufficiently. Good luck!

LIST 1: MOVIES

CRFMCBAN (6)  
KOZFTMVBOUS (8)  
LGHFNCE (4)  
LSRHWE (5)  
MRXX (4)  
VDOJION (4)  
VTZXRVOIR (4)  
YXDVHXNED (6)  


LIST 2: INFLUENTIAL PEOPLE

AFESCWOOD (5)  
DFHZPSGTON (6)  
DMUSFKSON (6)  
IZHZGE H W (4)  
MKFDWMA (7)  
QCSNDK (5)  
QJRFIAM HOWARD (4)  
UTVBEJY (6)  


LIST 3: SONGS

BFBSY (4)  
HTELYY (5)  
KDGAEGIAN (4)  
MSYYWTACK (6)  
OTAPQYE (5)  
RZCE (3)  
RZNYPOINE (6)  
WULCMOR (5)  


LIST 4: FICTIONAL CHARACTERS

EGOWEEDORE (5)  
LSWUWUS (5)  
OTOSLEY (3)  
OWSAPTFE (7)  
QSMIDY (5)  
XPLLDR (5)  
XZRHOR (3)  
ZTJPNAYTOM (7)  
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I notice straight away that

The items are only enciphered up to the number of letters in the smaller, hidden part. For example, in EGOWEEDORE (5), the first five letters of the word are enciphered and the rest are not.

This means that even without knowing the cipher, I was able to work out a lot of these right off the bat, and from that I was able to work out that the cipher is

Simply adding the letter values of the smallest word to those of the biggest word. To use the same example, EGOWEEDORE (5) is Albus Dumbledore:

D(4) + A(1) = E(5)
U(21) + L(12) = G(33-26=7)
M(13) + B(2) = O(15)
B(2) + U(21) = W(23)
L(12) + S(19) = E(31-26=5)

I won't write out the sums for every single item (at least, not right now), but here are the deciphered items:

Movies:

CRFMCBAN (6) = American Beauty
KOZFTMVBOUS (8) = Inglourious Basterds
LGHFNCE (4) = True Romance (credit to @Wa Kai)
LSRHWE (5) = Jackie Brown
MRXX (4) = Kill Bill
VDOJION (4) = Pulp Fiction
VTZXRVOIR (4) = Reservoir Dogs
YXDVHXNED (6) = Django Unchained

The odd one out is

American Beauty, as it wasn't directed by Quentin Tarantino

Influential People:

AFESCWOOD (5) = Frank Underwood
DFHZPSGTON (6) = George Washington
DMUSFKSON (6) = Thomas Jefferson
IZHZGE H W (4) = George H W Bush
MKFDWMA (7) = Abraham Lincoln
QCSNDK (5) = Barack Obama
QJRFIAM HOWARD (4) = William Howard Taft
UTVBEJY (6) = John F Kennedy

The odd one out is

Frank Underwood, who is a fictional character - the rest are all real US Presidents

Songs:

BFBSY (4) = Penny Lane
HTELYY (5) = Space Oddity (credit to @hexomino)
KDGAEGIAN (4) = Norwegian Wood
MSYYWTACK (6) = Paperback Writer
OTAPQYE (5) = Hello Goodbye
RZCE (3) = Hey Jude
RZNYPOINE (6) = Yellow Submarine
WULCMOR (5) = Eleanor Rigby

The odd one out is

Space Oddity because the rest are all Beatles songs

Fictional Characters:

EGOWEEDORE (5) = Albus Dumbledore
LSWUWUS (5) = Severus Snape
OTOSLEY (3) = Ron Weasley
OWSAPTFE (7) = Hermione Granger
QSMIDY (5) = Draco Malfoy
XPLLDR (5) = Harry Potter
XZRHOR (3) = Lex Luthor
ZTJPNAYTOM (7) = Neville Longbottom (credit to @Wa Kai)

Which makes the odd one out

Lex Luthor, as he's the only one not from Harry Potter

So we now have

American Beauty, Frank Underwood, Space Oddity, and Lex Luthor

I noticed that

Kevin Spacey was in American Beauty, played Frank Underwood in House of Cards, and played Lex Luthor in Superman Returns

Which means the odd one is

Space Oddity - it has 'space' in it, but no Kevin Spacey

It fits the theme of the puzzle because

Being the odd one out, it's an 'oddity'.

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  • 1
    $\begingroup$ The last missing movie must be 'True Romance' and 'Neville Longbottom' for the characters $\endgroup$ – Wa Kai Feb 3 '17 at 13:10
  • $\begingroup$ I would have worked out "True Romance" myself a few minutes ago had my Internet not crashed when I tried Googling Tarantino's filmography. Added those now. $\endgroup$ – F1Krazy Feb 3 '17 at 13:14
  • 1
    $\begingroup$ @F1Krazy Space Oddity? $\endgroup$ – hexomino Feb 3 '17 at 13:32
  • $\begingroup$ Well done, by the way, +1! $\endgroup$ – hexomino Feb 3 '17 at 13:33

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