Posted by **Jane** on Friday, November 23, 2007 at 5:09pm.

I would like to know if my answers are correct. I also need help solving some exercises the right way.

(a) Explain why [13] is invertible in

Z50.

Ans: it's invertible because

there's some [t]E Z50 such that

[13][t]=1

(b) Use the usual algorithm to find

[13]^(-1)and give your anwer

mod50.

Ans: [13]^(-1) = ? mod50

I wasn't sure how to use the usual

algorithm to solve this so I

used the matrix form, hopefully I

did it right. But I know that

I won't get full credit for it.

So, I would like to know the

right way of doing it.

__ __ __ __

| 1 0 | 50 | ~ | 1 -3 | 11 | ~

| 0 1 | 13 | | 0 1 | 13 |

-- -- -- --

__ __ ___ ___

| 1 -3 | 11 | ~| 6 -23 | -1 |

| -1 4 | 2 | | -1 4 | 2 |

-- -- -- --

1=6[50] -23[13] ==> [1]= -23[13]

==> [13]^(-1) = [-23]

(Can i just leave it like that or do I

have to continue?)

long division :

2 |So, 23+ remainder =

____ | 23 + 4= 27

23 | 50 |Therefore.

46 |

---- | [t]=[27] mod 50

4 | ===> [13][27]=1

(c) Use your answer in part (b) to solve the congruence 13x= 20mod 50.

and give your answer in mod50.

Ans: I wasn't sure how to solve it, so I did it the long way, one by one.

a =bmodn

13X=20mod 50 ==> 50 | 13x- 20

When x=40 ==>

50 | 13(40)-20 => it has

remainder 20 when

I do long division.

But I would like to know how to do it the right way.

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