can someone help please asap
1. Using the equation
(AB) mod C = ((Amod C)B) mod C
Solve for y in the following equation
A mod 7 = 2
A5 mod 7 =Y
Answer is y ==
2.
- Using Euclid’s formula, show the necessary steps to find thegreatest common divisor (gcd) of 342 and 243
gcd(a,b) == gcd(b,a modb)
gcd(243,342) == gcd(342,243 mod 342)
== gcd(342,99)
== gcd( )
Questions 2 3 4 use the followingsets that appeared In theEuclidGCDmodulus.doc document. We partition theintegers into three sets using x mod 3 (remainderis 0 1 or 2). The three sets are:
[0] = { … -9, -6, -3, 0 , 3 , 6 ,9 , 12, …. } can be written as 3k with k = …-2, -1, 0, 1, 2…
[1] = { … -8, -5, -2, 1 , 4 , 7 ,10, 13, …. } can be written as 3k+1 with k = …-2, -1, 0, 1, 2…
[2] = { … -7, -4, -1, 2 , 5 , 8 ,11, 14, …. } can be written as 3k +2 with k = …-2, -1 0, 1, 2…
Example:
-7 is in set [2] since -7 == 3(-3)+ 2
11 is in set [2] since 11 == 3(3)+ 2
-57 is in set [0] since -57==3(-19) + 0
71 is in set [2] since 71 == 3(23)+ 2
NOTE in the equation x = y modn x is always positive regardless of the value of y; n> 0
3.
- What is value of C in the following equation
C == -20 mod 7
Questions 2 3 4 use the followingsets that appeared In theEuclidGCDmodulus.doc document. We partition theintegers into three sets using x mod 3 (remainderis 0 1 or 2). The three sets are:
[0] = { … -9, -6, -3, 0 , 3 , 6 ,9 , 12, …. } can be written as 3k with k = …-2, -1, 0, 1, 2…
[1] = { … -8, -5, -2, 1 , 4 , 7 ,10, 13, …. } can be written as 3k+1 with k = …-2, -1, 0, 1, 2…
[2] = { … -7, -4, -1, 2 , 5 , 8 ,11, 14, …. } can be written as 3k +2 with k = …-2, -1 0, 1, 2…
Example:
-7 is in set [2] since -7 == 3(-3)+ 2
11 is in set [2] since 11 == 3(3)+ 2
-57 is in set [0] since -57==3(-19) + 0
71 is in set [2] since 71 == 3(23)+ 2
Expert Answer
Answer to can someone help please asap 1. Using the equation (AB) mod C = ((A mod C)B) mod C Solve for y in the following equation…