1 / 8

Number Theory

Number Theory. 04/26/07. Tasks. Announcement of Final Example lecture Course evaluation Dispatch computer project reports. Announcement of Final. The final is hold on Tuesday, May 8 th , 8:00am, Physics lecture hall Forty multiple choice problems Closed book.

hieu
Télécharger la présentation

Number Theory

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Number Theory 04/26/07

  2. Tasks • Announcement of Final • Example lecture • Course evaluation • Dispatch computer project reports

  3. Announcement of Final • The final is hold on Tuesday, May 8th, 8:00am, Physics lecture hall • Forty multiple choice problems • Closed book

  4. Example 1-modular exponentiation Eg.1: get 510 mod 21 Tip: just remember ab mod z=[(a mod z)(b mod z)] mod z Sol: Because 52 mod 21 =25 mod 21 =4, 510 mod 21= 45 mod 21= [(42 mod 21)(43 mod 21)] mod 21=[16*1]mod 21=16

  5. Example 2-solve ax+b mod z Eg.2: [10* 510 +11] mod 21 Tip: just remember [a+b] mod z=[(a mod z)+(b mod z)] mod z Sol: Because 510 mod 21 =16, 10*510 mod 21= 160 mod 21= 13 [10* 510 +11] mod 21=[13+11]mod 21=3

  6. Example 3-gcd Eg.3: gcd(2091,4807) Tip: just remember gcd(a,b)=gcd(b,r) Sol: gcd(2091,4807) = gcd(2091,625) = gcd(625,216) = gcd(216,193) = gcd(193,23) = gcd(23,9) =1 = gcd(9,5) = gcd(5,4) = gcd(4,1) = gcd(1,0)=1

  7. Example 4-inverse mod n Eg.4: We know gcd(3,4)=1, find inverse of 3 mod 4 Tip: Use Euclidean algorithm to find gcd(a,b)=sa+tb, and s mod b. Sol: gcd(3,4)=gcd(3,1)=gcd(1,0)=1 1=3-1*2; 1=4-1*3=4-(3-1*2)*3=4-3, so s=-1 s mod b=-1 mod 4=3

  8. Course Evaluation& dispatch reports • 14:332:202:01

More Related