Primitive Roots mod p - homepagesmathuicedu

Voorbeeld modulo 97 berekening: Voor factuur 325 van klant 1459 kan je bv de mededeling 000/1459/325XY aanmaken “XY” is daarbij dan het controlegetal en dat bereken je eenvoudigweg als volgt : Stap 1 Deel de eerste tien cijfers (0001459325) door 97; je bekomt 15 044,5876 Stap 2

MATH 3240Q Introduction to Number Theory Homework 7

order of 10 modulo 97 is 96, i e , ord 97(10) = 96 Hence, the length of the period is 96 5 282475249=19400 = 14560:5798453608247422680412371::: Here 19400 = 23 52 97 where 97 is prime so M= 3 and the repeating part of the decimal expansion starts with the M+1 = 4th digit, and the order of 10 modulo 97 is 96, i e , ord 97(10) = 96

Parts I and II of the Law of Quadratic Reciprocity

Example: (75/97) = (3•5•5/97) = (3/97)(5/97)(5/97) = (3/97)(±1)2 = (3/97) Since 102 = 100 ≡ 3 (mod 97), 3 is a QR modulo 97 Thus (75/97) = (3/97) = 1 Law of Quadratic Reciprocity Part I We took a prime p and found the QRs and NRs Now we are going to fix a and find which primes p have a as a QR

MATH 4440/5440 Assessment, Module 2 (Discrete Log Problem

(10 points) Compute the inverse of 30 modulo 97 by hand Note that 97 is prime Show your work neatly please, and show all steps Solution We use Gauss’ method First, divide 97 by 30: 97 = 3 30 + 7 We learn that 30 ( 3) 7 (mod 97): Then, divide 97 by 7: 97 = 13 7 + 6 Adding a copy of 7, we obtain 97 = 14 7 1 We learn that 7 14 1 (mod 97):

NOTA AIFA 97 allegato 1

Microsoft Word - NOTA AIFA 97 allegato 1 docx Author: domsc Created Date: 6/15/2020 4:06:31 PM

13 Congruences - NIU

stated modulo 90, and so the most satisfying answer is given in terms of congruence classes modulo 90 Solution: We have gcd(42,90) = 6, so there is a solution since 6 is a factor of 12 Solving the congruence 42x ≡ 12 (mod 90) is equivalent to solving the equation 42x= 12+90qfor integers xand q This reduces to 7x= 2+15q, or 7x≡ 2 (mod 15)