1) Using the Caser Cypher, decrypt KYVMR CLVFW KYVBV PZJJV MVEKV VE 3) Use publi
ID: 2902838 • Letter: 1
Question
1) Using the Caser Cypher, decrypt KYVMR CLVFW KYVBV PZJJV MVEKV VE
3) Use public key (k,m)=(3,2669) to decrypt 12151224147100230116. (hint: Use 00, 01, .... , 25 to denote a,b,....,z; and break the string into numbers of 4 digits)
1) Using the Caser Cypher, decrypt KYVMR CLVFW KYVBV PZJJV MVEKV VE 2) Decrypt the message RTOLKTOIK which was encrypted using the affine transformation C equiv3P+24 (mod 26) 3) Use public key (k,m)=(3,2669) to decrypt 12151224147100230116. (hint: Use 00, 01, .... , 25 to denote a,b,....,z; and break the string into numbers of 4 digits)Explanation / Answer
1. The message KYVMR CLVFW KYVBV PZJJV MVEKV VE was encrypted using
a shift transformation C ? P + k
The first thing to note is that the most frequently occurring letters in the ciphertext are
V, with eight occurrences, and K, with three. None of the other letters occur more than
twice. So it is reasonable to think that E is mapped to V and T is mapped to K under this
transformation. That is, (4 ? 21) and (19 ? 10). Plugging either of these into the given
equation C ? P +k (mod 26) gives a value of 17 for k. Using this we decrypt the ciphertext
to: THE VALUE OF THE KEY IS SEVENTEEN.
2. First note that the inverse of three modulo 26 is 9, as 3
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