10.4.32

# 10.4.32 - x ≡ 161 * 41 x ≡ 161 * 125 (mod 660) x ≡...

This preview shows page 1. Sign up to view the full content.

INFS 501 - 10.4.32 10.4.32 (a) 660 = 16*41+4, so 4 = 660 - 16*41, 41 = 10*4+1, so 1 = 41 - 10*4. Now substituting backward: 1 = 41 - 10*4 = 41 - 10*(660 - 16*41) 1 = -10 * 660 + 161 * 41 Thus, 161 ≡ 41 -1 (mod 660). 10.4.32 (b) Next, multiply 161 times both sides of the equation we want solve: 41 x ≡ 125 (mod 660)
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: x ≡ 161 * 41 x ≡ 161 * 125 (mod 660) x ≡ 20,125 ≡ 325+30*660 ≡ 325 (mod 660). 325 is the smallest (least) positive solution to 41 x ≡ 125 (mod 660). Check : 41*325 ≡ 13,325 ≡ 20*660 + 125 ≡ 125 (mod 660). Indeed, x=325 is a solution to 41x≡125 (mod 660) ....
View Full Document

## This note was uploaded on 11/01/2010 for the course INFS 501 taught by Professor Ellis,w during the Spring '08 term at George Mason.

Ask a homework question - tutors are online