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6.13 Two samples of RSA ciphertext are presented in Tables 6.2 and 6.3. Your task is to decrypt them. The public parameters of the system are n = 18923 and b = 1261 (for Table 6.2) and n = 31313 and b = 4913 (for Table 6.3). This can be accomplished as follows. First, factor n (which is easy because it is so small). Then compute the exponent a from po(n), and, finally, decrypt the ciphertext. Use the SQUARE-AND-MULTIPLY ALGORITHM to exponentiate modulo n. In order to translate the plaintext back into ordinary English text, you need to know how alphabetic characters are "encoded" as elements in Zn. Each element of Zn represents three alphabetic characters as in the following ex- amples: DOG 3 x 26² +14 × 26+6 = CAT → 2x 26² +0 x 26 +19 ZZZ = 25 x 26² +25 x 26+25= You will have to invert this process as the final step in your program. 2398 1371 17575.

6.13 Two samples of RSA ciphertext are presented in Tables 6.2 and 6.3. Your task is to decrypt them. The public parameters of the system are n = 18923 and b = 1261 (for Table 6.2) and n = 31313 and b = 4913 (for Table 6.3). This can be accomplished as follows. First, factor n (which is easy because it is so small). Then compute the exponent a from po(n), and, finally, decrypt the ciphertext. Use the SQUARE-AND-MULTIPLY ALGORITHM to exponentiate modulo n. In order to translate the plaintext back into ordinary English text, you need to know how alphabetic characters are "encoded" as elements in Zn. Each element of Zn represents three alphabetic characters as in the following ex- amples: DOG 3 x 26² +14 × 26+6 = CAT → 2x 26² +0 x 26 +19 ZZZ = 25 x 26² +25 x 26+25= You will have to invert this process as the final step in your program. 2398 1371 17575.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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6.13 Two samples of RSA ciphertext are presented in Tables 6.2 and 6.3. Your task is to decrypt them. The public parameters of the system are \( n = 18923 \) and \( b = 1261 \) (for Table 6.2) and \( n = 31313 \) and \( b = 4913 \) (for Table 6.3). This can be accomplished as follows. First, factor \( n \) (which is easy because it is so small). Then compute the exponent \( a \) from \( \phi(n) \), and, finally, decrypt the ciphertext. Use the SQUARE-AND-MULTIPLY ALGORITHM to exponentiate modulo \( n \). In order to translate the plaintext back into ordinary English text, you need to know how alphabetic characters are “encoded” as elements in \( Z_n \). Each element of \( Z_n \) represents three alphabetic characters as in the following examples: - \( \text{DOG} \rightarrow 3 \times 26^2 + 14 \times 26 + 6 = 2398 \) - \( \text{CAT} \rightarrow 2 \times 26^2 + 0 \times 26 + 19 = 1371 \) - \( \text{ZZZ} \rightarrow 25 \times 26^2 + 25 \times 26 + 25 = 17575 \) You will have to invert this process as the final step in your program.
Transcribed Image Text:6.13 Two samples of RSA ciphertext are presented in Tables 6.2 and 6.3. Your task is to decrypt them. The public parameters of the system are \( n = 18923 \) and \( b = 1261 \) (for Table 6.2) and \( n = 31313 \) and \( b = 4913 \) (for Table 6.3). This can be accomplished as follows. First, factor \( n \) (which is easy because it is so small). Then compute the exponent \( a \) from \( \phi(n) \), and, finally, decrypt the ciphertext. Use the SQUARE-AND-MULTIPLY ALGORITHM to exponentiate modulo \( n \). In order to translate the plaintext back into ordinary English text, you need to know how alphabetic characters are “encoded” as elements in \( Z_n \). Each element of \( Z_n \) represents three alphabetic characters as in the following examples: - \( \text{DOG} \rightarrow 3 \times 26^2 + 14 \times 26 + 6 = 2398 \) - \( \text{CAT} \rightarrow 2 \times 26^2 + 0 \times 26 + 19 = 1371 \) - \( \text{ZZZ} \rightarrow 25 \times 26^2 + 25 \times 26 + 25 = 17575 \) You will have to invert this process as the final step in your program.
In 6-13, decrypt the first 5 blocks: 12423–11524–7243–7459–14303 n = 18923, e = 1261
Transcribed Image Text:In 6-13, decrypt the first 5 blocks: 12423–11524–7243–7459–14303 n = 18923, e = 1261
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