## Compute the following for the given Z(n)?

**Question**

Write your answer in the from [a] where 0 is greater or = to a, and a is smaller or = to n-1.

1) [21] + [19] in Z(12)

2) [34][27] in Z(45)

3) [33]([83]-[67]) in Z(100)

4) The additive inverse of [18] in Z(26)

**Answer**

We use the fact, by definition,

[m] = 0 in Z(n) where,

for all m where m is divisible by n .

and [m] = “remainder of m when divided by n”.

1) [21] + [19] in Z(12)

So, in Z(12),

[21] + [19] = [21 + 19]

[21] + [19] = [40]

[21] + [19] = [12*3 + 4]

[21] + [19] = [4]

[21] + [19] = 4

2) [34][27] in Z(45)

[34][27] = [34*27]

[34][27] = [3427]

[34][27] = [3420 + 7]

[34][27] = [45*76 + 7]

[34][27] = [7]

[34][27] = 7

3) [33]([83]-[67]) in Z(100)

[33]([83]-[67]) = [33]([83 – 67])

[33]([83]-[67]) = [33] ([16])

[33]([83]-[67]) = [33*16]

[33]([83]-[67]) = [528]

[33]([83]-[67]) = [5*100 + 28]

[33]([83]-[67]) = [28]

[33]([83]-[67]) = 28

4) The additive inverse of [18] in Z(26)

Let the additive inverse be y.

By definition [y] + [18] = 26 in Z(26)

So,

[y] = 26 – [18]

[y] = [26] – [18]

[y] = [26 – 18]

[y] = [8]

Hence, y = 8.

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