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


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
up vote 4 down vote accepted

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.)

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