With the Euclidean algorithm we finally have an efficient algorithm for finding the multiplicative inverse in Zm that is much better than exhaustive search. Find the inverses in Zm of the following elements a modulo m: 1. a=7, m=26 (affine cipher) 2. a=19, m=999

a) The inverse of 7 in is 15

b) The inverse of 19 in is 631

Step-by-step explanation:

1)

26-3*7 = 5

7-5 = 2

5 - 2*2 = 1

Thus 1 = 5 - (2*(7-5) ) = 3 * 5 - 2*7 = 3 * (26-3*7) - 2*7 = 3*26 -11*7

Now, lets check 7 * (-11) = -77, and -63 + 3*26 = -77+78 = 1. We take as inverse -11+26 = 15 (so that it lies between 1 and 26).

2)

999/19 = 52.278....

999 - 52*19 = 11

19 - 11 = 8

11 - 8=3

8 - 2*3 = 2

3 - 2 = 1

Thus,

1 = 3-2 = 3 - (8-2*3) = 3*3-8 = 3*(11-8) - 8 = 3*11 - 4*8 = 3*11 - 4*(19-11) = 7*11-4*19 = 7*(999-52*19) - 4*19 = 7*999 - 368 * 19  

We take as inverse -368+999 = 631.

Lets check, -368*19 + 7 *999 = -6992 + 6993 = 1, it works.

i dont get the question

step-by-step explanation:

there is only one way to arrange all the eight relative to each other because they are not distinct. we have

now there are nine places to place an (and not more than one in one place).   because we have four symbols, we need to choose four of these nine places to insert an . order does not matter because the are not distinct.

thus there are 1 · c(9,4) ways, or, if we calculate it,

thus there are 126 ways to form such "words."