Everything You Need to Know About Modular Arithmetic
modulo 10 while 5 is not relatively prime to 10 and therefore has no inverse modulo 10 Ex 5 We can compute which numbers will have inverses modulo 10 by computing which are relatively prime to 10 = 5·2 These numbers are x = 1,3,7,9 It is easy to see that the following table gives inverses module 10: 2
Math 110 Homework 4 Solutions
powers of agenerate all units modulo p The primitive roots are 2;6;7;8 (mod 11) To check, we can simply compute the rst ˚(11) = 10 powers of each unit modulo 11, and check whether or not all units appear on the list A more sophisticated approach: Once you have a primitive root a(mod 11), it’s a fact that the other primitive roots must be
EECS 203-1 Homework 10 Solutions Total Points: 30
14) a) Show that the positive integers less than 11, except 1 and 10, can be split into pairs of integers such that each pair consists of integers that are inverses of each other modulo 11 b) Use part (a) to show that 10 -1 (mod 11) 6 points a) We know that each of the integers 1, 2, 3 10 has an inverse modulo 11 (because 11 is prime)
Math 110 Problem Set 2 Solutions
root mod 11 (b) Note that the inverse of 3 (mod 10) is 7 (mod 10) Now, by Fermat’s Little Theorem, 210 · 1 (mod 11) Thus, 23 · 8 (mod 11))(23)7 · 87 (mod 11))87 · 221 · 2 (mod 11) Thus, the x we want is 7 (c) Let c be a congruence class mod 11 By (a), there exists x such that 2x · c (mod 11) Applying the formula from (b), we have
CHECK DIGITS - An application of MODULAR ARITHMETIC
10 is the check digit 0 10 4 9 7 8 1 7 3 6 1 5 0 4 5 3 5 2 7 1 { 0 mod 11 ISBN 0-471-31055-7 This detects all single digit errors and all transpositions of adjacent digits The only drawback is that the “check digit” could be “10” which is not a single digit In this case one uses “X”: 0 10 4 9 7
Socioemocional - Proyecto Alcanza
3 INTRODUCCIÓN Annette López de Méndez, Ed D Directora Proyecto ALCANZA Para crecer bien, es de suma importancia atender los aspectos físico, cognitivo y emocional
13 Congruences - NIU
39 Prove that 10n+1 +4 ·10n +4 is divisible by 9, for all positive integers n Comment: This could be proved by induction, but we can give a more elegant proof using congruences Solution: The proof consists of simply observing that 10n+1+4·10n+4 ≡ 0 (mod 9) since 10 ≡ 1 (mod 9) 40 Prove that for any integer n, the number n3+5nis
Fermat’s Little Theorem Solutions - CMU
Sep 27, 2015 · x103 4 mod 11 [Solution: x 5 mod 11] By Fermat’s Little Theorem, x10 1 mod 11 Thus, x103 x3 mod 11 So, we only need to solve x3 4 mod 11 If we try all the values from x = 1 through x = 10, we nd that 53 4 mod 11 Thus, x 5 mod 11 9 Find all integers x such that x86 6 mod 29 [Solution: x 8;21 mod 29] By Fermat’s Little Theorem, x28 1
MÓDULO PARA REMEDIAR - Departamento de Educación
c 11 d 10 Estándar: Conservación y Cambio Expectativa ES BCB3 CC2: Aplica conceptos estadísticos y de probabilidad para explicar la variación y distribución de las características visibles en la población El énfasis está en el uso de las matemáticas para describir la
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