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The goal of this exercise is to work thru the RSA system in a simple case: We will use primes p = 53, q = 41 and form n = 53 · 41 = 2173. (This is typical of the RSA system which chooses two large primes at random generally, and multiplies them to find n. The public will know n but p and q will be kept private.] Now we choose our public key e = 11. This will work since gcd(11, (p – 1)(q – 1)) = gcd(11, 2080) = 1. [In general as long as we choose an 'e' with gcd(e,(p-1)(q-1))=1, the system will work.] Next we encode letters of the alphabet numerically say via the usual: (A=0,B=1,C=2,D=3,E=4,F=5,G=6,H=7,l=8, J=9,K=10,L=11,M=12,N=13,0=14,P=15,Q=16,R=17, S=18,T=19,U=20,V=21,W=22,X=23,Y=24,Z=25.) We will practice the RSA encryption on the single integer 15. (which is the numerical representation for the letter "P"). In the language of the book, M=15 is our original message. The coded integer is formed via c = Me mod n. Thus we need to calculate 15' mod 2173. This is not as hard as it seems and you might consider using fast modular multiplication. The canonical representative of 15' mod 2173 is

The goal of this exercise is to work thru the RSA system in a simple case: We will use primes p = 53, q = 41 and form n = 53 · 41 = 2173. (This is typical of the RSA system which chooses two large primes at random generally, and multiplies them to find n. The public will know n but p and q will be kept private.] Now we choose our public key e = 11. This will work since gcd(11, (p – 1)(q – 1)) = gcd(11, 2080) = 1. [In general as long as we choose an 'e' with gcd(e,(p-1)(q-1))=1, the system will work.] Next we encode letters of the alphabet numerically say via the usual: (A=0,B=1,C=2,D=3,E=4,F=5,G=6,H=7,l=8, J=9,K=10,L=11,M=12,N=13,0=14,P=15,Q=16,R=17, S=18,T=19,U=20,V=21,W=22,X=23,Y=24,Z=25.) We will practice the RSA encryption on the single integer 15. (which is the numerical representation for the letter "P"). In the language of the book, M=15 is our original message. The coded integer is formed via c = Me mod n. Thus we need to calculate 15' mod 2173. This is not as hard as it seems and you might consider using fast modular multiplication. The canonical representative of 15' mod 2173 is

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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The goal of this exercise is to work thru the RSA system in a simple case: We will use primes p = 53, q = 41 and form n = 53 · 41 = 2173. [This is typical of the RSA system which chooses two large primes at random generally, and multiplies them to find n. The public will know n but p and q will be kept private.] Now we choose our public key e = 11. This will work since gcd(11, (p – 1)(g – 1)) = gcd(11, 2080) = 1. [In general as long as we choose an 'e' with gcd(e,(p-1)(q-1))=1, the system will work.] Next we encode letters of the alphabet numerically say via the usual: (A=0,B=1,C=2,D=3,E3D4,F=5,G36,H=7,l=8, J=9,K=10,L=11,M=12,N=13,0=14,P=15,Q=16,R=17, S=18,T=19,U=20,V=21,W=22,X=23,Y=24,Z=25.) We will practice the RSA encryption on the single integer 15. (which is the numerical representation for the letter "P"). In the language of the book, M=15 is our original message. The coded integer is formed via c = Me mod n. Thus we need to calculate 15' mod 2173. This is not as hard as it seems and you might consider using fast modular multiplication. The canonical representative of 15' mod 2173 is
Transcribed Image Text:The goal of this exercise is to work thru the RSA system in a simple case: We will use primes p = 53, q = 41 and form n = 53 · 41 = 2173. [This is typical of the RSA system which chooses two large primes at random generally, and multiplies them to find n. The public will know n but p and q will be kept private.] Now we choose our public key e = 11. This will work since gcd(11, (p – 1)(g – 1)) = gcd(11, 2080) = 1. [In general as long as we choose an 'e' with gcd(e,(p-1)(q-1))=1, the system will work.] Next we encode letters of the alphabet numerically say via the usual: (A=0,B=1,C=2,D=3,E3D4,F=5,G36,H=7,l=8, J=9,K=10,L=11,M=12,N=13,0=14,P=15,Q=16,R=17, S=18,T=19,U=20,V=21,W=22,X=23,Y=24,Z=25.) We will practice the RSA encryption on the single integer 15. (which is the numerical representation for the letter "P"). In the language of the book, M=15 is our original message. The coded integer is formed via c = Me mod n. Thus we need to calculate 15' mod 2173. This is not as hard as it seems and you might consider using fast modular multiplication. The canonical representative of 15' mod 2173 is
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