Cryptographic Coding

If the word “BRIGHT” is coded as” OCPLKV”, then how will you code the word “SERIAL”?

1) CDKYFG
3) CKDGFY
2) FPYNDN
4)FPNDYN

5) none of these Can someone give me the answer with a little explanation?

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Are you told what method of encryption is used? –  Michael Albanese Sep 13 '12 at 10:52
@MichaelAlbanese Not exactly but I know that we do not use the make the same substitution all the time, there is some algorithm running. –  user221287 Sep 13 '12 at 10:58
Suppose it is a simple substitution cipher. Then, R and I are coded as C and P respectively. R and I also occur in SERIAL (in fact the doublet RI occurs in both BRIGHT and SERIAL). Are any of the four suggested answers a possibility? –  Dilip Sarwate Sep 13 '12 at 11:01
@DilipSarwate, that only works if you make the same substitutions all the time. –  mikeazo Sep 13 '12 at 11:03
"That only works..." Absolutely true, and without any further information about the method of encryption, the problem has no answer: SERIAL could be encoded into any sequence of six letters (including the four listed, and so "None of the above" is also not a correct answer to the question). –  Dilip Sarwate Sep 13 '12 at 11:08

If you map the letters of the words "BRIGHT" and "OCPLKV" to numbers using the obvious encoding A = 1, B = 2, ..., Z = 26, you get $$2, 18, 9, 7, 8, 20$$ and $$15, 3, 16, 12, 11, 22.$$ If you now subtract the former sequence from the latter, what's left is $$13, -15, 7, 5, 3, 2.$$ Assuming that the cipher is using arithmetic modulo 26, the −15 becomes 26 − 15 = 11, leaving us with the sequence $$13, 11, 7, 5, 3, 2.$$ Does that look familiar? What do you get when you add these numbers modulo 26 to the letters of the word "SERIAL", encoded into numbers as above? (Hint: it's one of the four choices.)