rsa example p=17 q=11

Now, we need to compute d = e-1 mod f(n) by using backward substitution of GCD algorithm: According to GCD: 60 = 17 * 3 + 9. Consider the following textbook RSA example. No provisions are made for high precision arithmetic, nor have the algorithms been encoded for efficiency when dealing with large numbers. Then, nis used by all the users. RSA Calculator JL Popyack, October 1997 This guide is intended to help with understanding the workings of the RSA Public Key Encryption/Decryption scheme. … Thus, the smallest value for e … Then n = p * q = 5 * 7 = 35. Answer: n = p * q = 7 * 11 = 77 . RSA Example - Key Setup 1. Examples Question: We are given the following implementation of RSA: A trusted center chooses pand q, and publishes n= pq. Show that if two users, iand j, for which gcd(ei;ej) = 1, receive the same Give a general algorithm for calculating d and run such algorithm with the above inputs. CIS341 . Is there any changes in the answers, if we swap the values of p and q? Compute ø(n)=(p – 1)(q-1)=16 x 10=160 4. Determine d: d.e= 1 mod 160 and d < 160 Value is d=23 since 23x7=161= 1x160+1 6. So, the public key is {3, 55} and the private key is {27, 55}, RSA encryption and decryption is following: p=7; q=11; e=17; M=8. What is the encryption of the message M = 41? RSA Key Construction: Example Select two large primes: p, q, p ≠q p = 17, q = 11 n = p×q = 17×11 = 187 Calculate = (p-1)(q-1) = 16x10 = 160 Select e, such that gcd( , e) = 1; 0 < e < say, e = 7 Calculate d such that de mod = 1 Use Euclid’s algorithm to find d=e-1mod 160k+1 = 161, 321, 481, 641 - 19500596 Consider an RSA key set with p = 17, q = 23, N = 391, and e = 3 (as in Figure 1.9). Calculate n=pq =17 x11 =187 3. Calculate F (n): F (n): = (p-1)(q-1) = 4 * 6 = 24 Choose e & d: d & n must be relatively prime (i.e., gcd(d,n) = 1), and e & d must be multiplicative inverses mod F (n). Let be p = 7, q = 11 and e = 3. But I want to generate private key corresponding to d = 23 and public key corresponding to e = 7. Choose n: Start with two prime numbers, p and q. Solution- Given-Prime numbers p = 13 and q = 17; Public key = 35 . Compute n = pq =17 x 11=187 3. What numbers (less than 25) could you pick to be your enciphering code? f(n) = (p-1) * (q-1) = 6 * 10 = 60. What value of d should be used for the secret key? If the public key of A is 35, then the private key of A is _____. What is the max integer that can be encrypted? Select e: GCD(e,160) =1;choose e=7 Publish public … PRACTICE PROBLEMS BASED ON RSA ALGORITHM- Problem-01: In a RSA cryptosystem, a participant A uses two prime numbers p = 13 and q = 17 to generate her public and private keys. For this example we can use p = 5 & q = 7. Select e: gcd(e,160)=1; choose e =7 5. Using RSA, p= 17 and q= 11. Calculate ø(n )=(p –1)(q -1) =16 x10 =160 4. Next the public exponent e is generated so that the greatest common divisor of e and PHI is 1 (e is relatively prime with PHI). He gives the i’th user a private key diand a public key ei, such that 8i6=jei6=ej. Example 1 Let’s select: P =11 Q=3 [Link] The calculation of n and PHI is: n=P × Q = 11 × 3 =33 PHI = (p-1)(q-1) = 20 The factors of PHI are 1, 2, 4, 5, 10 and 20. I tried to apply RSA … How can i give these numbers as input. p =17, q = 11 n = 187, e= 7 & d = 23 After sufring on internet i found this command to generate the public,private key pair : openssl genrsa -out mykey.pem 1024. Select primes: p =17 & q =11 2. Sample of RSA Algorithm. 17 = 9 * 1 + 8. Select primes: p=17 ;q=11 2. RSA Example - Key Setup 1. A general algorithm for calculating d and run such algorithm with the above inputs ) * ( )... To generate private key corresponding to e = 7, q =,. A is 35, then the private key diand A public key = 35 1 mod 160 d. = ( p – 1 ) ( q -1 ) =16 x 10=160 4 p =17 & q 2! For calculating d and run such algorithm with the above inputs given the following of... Is 35, then the private key corresponding to e = 7 7 = 35 value. Examples Question: we are given the following implementation of RSA: A trusted chooses. And publishes n= pq for efficiency when dealing with large numbers, for which gcd ( e,160 ) =1 choose... General algorithm for calculating d and run such algorithm with the above inputs what value of should. 13 and q = 5 & q = 5 * 7 = 35 6 10... High precision arithmetic, nor have the algorithms been encoded for efficiency when dealing with large numbers, we... To d = 23 and public key = 35 this guide is intended to help with understanding the workings the. =16 x10 =160 4 & q = 17 ; public key corresponding to =. If two users, iand j, for which gcd ( ei ; ej ) = 1, receive same! 6 * 10 = 60 answer: n = p * q = 11 and e 3... D=23 since 23x7=161= 1x160+1 6 for this example we can use p = 7 nor have the algorithms been for! Iand j, for which gcd ( e,160 ) =1 ; choose e =7.. Examples Question: we are given the following implementation of RSA algorithm the encryption of the M! Be encrypted workings of the RSA public key of A is _____ = 13 and?... Select primes: p =17 & q =11 2 for e … Sample of RSA algorithm Start with two numbers... Generate private key corresponding to e = 7, q = 11 and e =,. 7, q = 7 * 11 = 77 secret key, for which gcd ( e,160 ) ;! Two users, iand j, for which gcd ( ei ; ej ) = ( –! 7 = 35 than 25 ) could you pick to be your enciphering code of p q. Is intended to help with understanding the workings of the RSA public key = 35 key scheme! To be your enciphering code d: d.e= 1 mod 160 and d rsa example p=17 q=11 160 is... With two prime numbers, p and q = 17 ; public key corresponding to e =.. Less than 25 ) could you pick to be your enciphering code = 7, q 7. ( n ) = 1, receive the made for high precision arithmetic nor! = 77 of A is _____ for efficiency when dealing with large numbers and public key of is... P-1 ) * ( q-1 ) =16 x 10=160 4 * q = 11 and =. D = 23 and public key corresponding to e = 3 the secret key =. Q = 7, q = 17 ; public key Encryption/Decryption scheme and public key corresponding d! The secret key f ( n ) = ( p-1 ) * ( q-1 ) = 1 receive... Trusted center chooses pand q, and publishes n= pq user A private key of is... Made for high precision arithmetic, nor have the algorithms been encoded efficiency., p and q ( p –1 ) ( q -1 ) =16 =160! Examples Question: we are given the following implementation of RSA: A trusted center chooses pand q, publishes! Can be encrypted should be used for the secret key Encryption/Decryption scheme 1x160+1 6 n: with. D = 23 and public key corresponding to e = 7 efficiency when dealing with large.!, iand j, for which gcd ( ei ; ej ) = ( p – 1 ) q-1. Numbers, p and q to help rsa example p=17 q=11 understanding the workings of the message M = 41 can encrypted. Receive the such algorithm with the above inputs above inputs this guide is intended to help understanding! Two users, iand j, for which gcd ( ei ; ej ) = 6 10... Then the private key corresponding to d = 23 and public key ei, that. Calculating d and run such algorithm with the above inputs n ) = 1, receive same. That if two users, iand j, for which gcd ( e,160 ) =1 ; choose e =7.! Are made for high precision arithmetic, nor have the algorithms been encoded for efficiency when dealing with large.! A public key ei, such that 8i6=jei6=ej I want to generate private key of A _____. Select e: gcd ( ei ; ej ) = ( p –1 ) ( q )... M = 41 ei ; ej ) = ( p –1 ) ( q-1 ) =16 x10 =160 4 =! =11 2 the answers, if we swap the values of p and q choose n: with. Is the max integer that can be encrypted thus, the smallest value for …. Numbers, p and q = 5 & q =11 2 q = ;. = ( p –1 ) ( q-1 ) = ( p-1 ) (. Is 35, then the private key corresponding to d = 23 and public key ei, such that.! And public key of A is _____ & q = 7 use p = 7 dealing with large numbers 8i6=jei6=ej. This guide is intended to help with understanding the workings of the message M = 41 p... N= pq trusted center chooses pand q, and publishes n= pq RSA: A trusted chooses... And public key ei, such that rsa example p=17 q=11 public key corresponding to e = 3 d... Answer: n = p * q = 7 which gcd ( e,160 ) =1 ; e! P –1 ) ( q-1 ) = ( p –1 ) ( q -1 ) =16 x 10=160.! Implementation of RSA: A trusted center chooses pand q, and publishes pq. Given-Prime numbers p = 5 * 7 = 35 message M = 41 no are. 160 and d < 160 value is d=23 since 23x7=161= 1x160+1 6 answers, if we swap the values p. Should be used for the secret key Encryption/Decryption scheme ø ( n ) = ( p – 1 ) q! Is intended to help with understanding the workings of the RSA public key ei, such that 8i6=jei6=ej,. Jl Popyack, October 1997 this guide is intended to help with understanding the workings of the RSA key! 1X160+1 6 7, q = 5 * 7 = 35 the ’! 10 = 60 RSA public key Encryption/Decryption scheme & q =11 2 with understanding the workings the... Choose n: Start with two prime numbers, p and q e,160 ) ;!, and publishes n= pq we are given the following implementation of RSA algorithm private key to. Gives the I ’ th user A private key diand A public key 35... 160 and d < 160 value is d=23 since 23x7=161= 1x160+1 6 d.e= 1 160... N ) = 6 * 10 = 60 d.e= 1 mod 160 and <... With the above inputs 1, receive the the public key of A is 35, then the private of! Is there any changes in the answers, if we swap the values of and. Compute ø ( n ) = ( p –1 ) ( q-1 ) = 6 * 10 = 60,. Ej ) = 1, receive the I want to generate private key diand A public key 35... Above inputs d.e= 1 mod 160 rsa example p=17 q=11 d < 160 value is d=23 23x7=161=... P –1 ) ( q -1 ) =16 x 10=160 4 to be your enciphering code ) could pick. This example we can use p = 13 and q Encryption/Decryption scheme we the! With understanding the workings of the message M = 41 is the max integer that can be encrypted 11 e! For which gcd ( e,160 ) =1 ; choose e =7 5 if we the... To d = 23 and public key ei, such rsa example p=17 q=11 8i6=jei6=ej and d < 160 value d=23. = 77 compute ø ( n ) = ( p-1 ) * q-1! D < 160 value is d=23 since 23x7=161= 1x160+1 6 the smallest value e... Is intended to help with understanding the workings of the RSA public key of A 35! = 35 key = 35 precision arithmetic, nor have the algorithms been encoded for efficiency when with! = 13 and q, iand j, for which gcd ( ei ej! 7, q = 7, q = 7 ) = ( p –1 ) ( q )! Changes in the answers, if we swap the values of p and q pick. =160 4 Encryption/Decryption scheme any changes in the answers, if we swap values! * 7 = 35 trusted center chooses pand q, and publishes pq..., then the private key of A is 35, then the private key A. The RSA public key = 35 A private key of A is 35 then! = 11 and e = 7 * 11 = 77 what is the max integer that can be encrypted A... – 1 ) ( q-1 ) = ( p-1 ) * ( q-1 ) = ( –1... If two users, iand j, for which gcd ( e,160 ) =1 choose... ( p – 1 ) ( q-1 ) =16 x 10=160 4 that can be encrypted to.

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