Midterm_2_2008_sol_Slides

Midterm_2_2008_sol_Slides - COMP 170 – Fall 2008 Midterm...

Info iconThis preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: COMP 170 – Fall 2008 Midterm 2 Solution Q1. Bob is constructing an RSA key-pair. He first chooses p = 11 , q = 19 and sets n = 11 · 19 = 209 . He then constructs his public key e and private key d and publishes the ( n, e ) pair. (a) Bob’s private key is d = 7 . What is the value of his public key e ? . Q1. Bob is constructing an RSA key-pair. He first chooses p = 11 , q = 19 and sets n = 11 · 19 = 209 . He then constructs his public key e and private key d and publishes the ( n, e ) pair. (a) Bob’s private key is d = 7 . What is the value of his public key e ? . By the definition of the RSA algorithm d · e mod T = 1 where T = ( p- 1)( q- 1) = 10 · 18 = 180 . Using, e.g., the extended GCD algorithm, we find that the multiplicative inverse of 7 mod T is e = 103 . Q1. Bob is constructing an RSA key-pair. He first chooses p = 11 , q = 19 and sets n = 11 · 19 = 209 . He then constructs his public key e and private key d and publishes the ( n, e ) pair. (b) Alice wants to send Bob a message M, < M < n. She calculates X = M e mod n to send Bob and finds that X = 15 . What is the value of the original message M ? Q1. Bob is constructing an RSA key-pair. He first chooses p = 11 , q = 19 and sets n = 11 · 19 = 209 . He then constructs his public key e and private key d and publishes the ( n, e ) pair. (b) Alice wants to send Bob a message M, < M < n. She calculates X = M e mod n to send Bob and finds that X = 15 . What is the value of the original message M ? M = X d mod n = 15 7 mod 209 = 203 . (The last equality can be derived any of multiple ways) Q2(a) Is (15 60 mod 61) = (15 62 mod 63)? Q2(b) Is (100 440 mod 89) = (100 1320 mod 89)? Q2(c) Evaluate 3 1052 mod 60 . Q2.(a) Is ( 15 60 mod 61 ) = ( 15 62 mod 63 ) ? No. 61 is a prime number so, by Fermat’s little theorem, 15 60 mod 61 = 1 . ....
View Full Document

This note was uploaded on 08/25/2010 for the course COMP COMP170 taught by Professor M.j.golin during the Spring '10 term at HKUST.

Page1 / 29

Midterm_2_2008_sol_Slides - COMP 170 – Fall 2008 Midterm...

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online