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CF CA AH AJ BA AG CF CA CH CJ AG

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  • $\begingroup$ Added "in" and "out". Hopefully it makes more sense now. This was the last edit on the puzzle. $\endgroup$ May 24, 2022 at 8:28

1 Answer 1

7
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The answer hidden in this puzzle is:

FAMOUS FACES

First, let's identify all of the people depicted here (using reverse image search where necessary - some of these are very obscure):

A = Neo (Keanu Reeves' character from The Matrix movie series)
B = Tom Cruise
C = Eros Ramazzotti
D = Harrison Ford
E = Steve Martin
F = Rix Robinson
G = Nina Simone
H = Zack Heart
I = Edith Piaf
J = Vien Hong

Now, look at the red number in brackets in the top-left corner of each image. For each of these people...

...either their first or last name (as per the title) is of this length:

A (3) = Neo
B (3) = Tom Cruise
C (4) = Eros Ramazzotti
D (4) = Harrison Ford
E (5) = Steve Martin
F (3) = Rix Robinson
G (4) = Nina Simone
H (5) = Zack Heart
I (5) = Edith Piaf
J (4) = Vien Hong (ambiguous at this point)

(i.e. the red symbol in the top 'instruction' image indicates that red numbers represent the length of a word)

Now look at the blue numbers in the bottom left corners (present in all but the first image)...

These are always less than or equal to the length of the word in question, suggesting they hint at a particular letter within the word. In fact, if we look at these indicated letters, in every instance the letter in this position can be altered so that the name is an anagram of a single-digit number spelled out. (NB No letter is indicated for 'Neo' since this is already an anagram of 'ONE'...)

A (3) [-] = NeoONE
B (3) [3] = To[m] Cruise → T[W]O
C (4) [4] = Ero[s] Ramazzotti → [Z]ERO
D (4) [4] = Harrison For[d]FO[U]R
E (5) [2] = S[t]eve Martin → SEVE[N]
F (3) [1] = [R]ix Robinson → [S]IX
G (4) [4] = Nin[a] Simone → NIN[E]
H (5) [3] = Zack He[a]rtTHRE[E]
I (5) [2] = E[d]ith Piaf → EI[G]HT
J (4) [4] = Vie[n] Hong → [F]IVE

(i.e. the blue 'Out' arrow in the top 'instruction' image indicates that blue numbers point to a particular letter that needs to be substituted out; the green 'In' arrow just suggests that something else needs to be subbed in...)

All that remains is for us to extract a final answer using the key: CF CA AH AJ BA AG CF CA CH CJ AG...

To do this, for each image A-J consider the corresponding anagrammed number 0-9. Then substitute these numbers into the key-phrase, replacing their corresponding letters. So every 'A' should become '1', every 'B should become '2', then C=0, D=4, E=7, F=6, G=9, H=3, I=8 and J=5.

Our key-phrase then transforms into:
06 01 13 15 21 19 06 01 03 05 19

All of these values fall in the range 1-26, so one step remains: Convert these values via A1Z26 to their corresponding letters of the alphabet. The result is FAMOUS FACES and the puzzle is solved!

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  • $\begingroup$ @JLee It took me a while of staring to spot the pattern! I think the thing at the top needs to be read like this: rot13(erq = jbeq yratgu, oyhr = fjnc bhg guvf yrggre, naq fb cerfhznoyl terra vf jung jr fjnccrq va.) But I can't quite spot the final extraction yet... $\endgroup$
    – Stiv
    May 24, 2022 at 22:42
  • $\begingroup$ @Stiv The last step: rot13(Qba'g bireguvax vg. Lbh'ir tbg yrggref sebz N-W gung znc gb qvtvgf sebz 0-9... whfg znc gurz, naq gura qb fbzrguvat cerggl fvzcyr jvgu gur erfhyg.) $\endgroup$
    – SQLnoob
    May 24, 2022 at 23:35
  • $\begingroup$ It appears to be moving in the same way for each one (+). I don’t want to put anymore details in case I spoil the answer. $\endgroup$
    – user80007
    May 24, 2022 at 23:47
  • $\begingroup$ Great job! You got the hardest part. For the last part, look at what @SQLnoob said. $\endgroup$ May 25, 2022 at 7:17
  • $\begingroup$ @SQLnoob Thanks for the prod! Sometimes it's incredibly hard to undo the way you've previously been thinking and start again from scratch. Strangely this phrase was one of the candidates I'd had for the final answer because of the letter repeat positions in the key, but I just could not spot how to make 'CF' equate to an 'F' from letter extractions from the names. Now I know why! $\endgroup$
    – Stiv
    May 25, 2022 at 8:48

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