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## Compute additive orders in Z/(n)

Find the orders of the following elements of the additive group $\mathbb{Z}/(36)$: $\overline{1}$, $\overline{2}$, $\overline{6}$, $\overline{9}$, $\overline{10}$, $\overline{12}$, $\overline{-1}$, $\overline{-10}$, $\overline{-18}$.

Solution:

$\overline{n}$ Reasoning $\overline{n}$
$\overline{1}$ 36 is the smallest multiple of 1 that is congruent to 0 mod 36. 36
$\overline{2}$ 36 is the smallest multiple of 2 that is congruent to 0 mod 36. 18
$\overline{6}$ Multiples of $\overline{6}$ are $\overline{6}$, $\overline{12}$, $\overline{18}$, $\overline{24}$, $\overline{30}$, $\overline{36} = \overline{0}$ 6
$\overline{9}$ Multiples of $\overline{9}$ are $\overline{9}$, $\overline{18}$, $\overline{27}$, $\overline{36} = \overline{0}$ 4
$\overline{10}$ Multiples of $\overline{10}$ are $\overline{10}$, $\overline{20}$, $\overline{30}$, $\overline{4}$, $\overline{14}$, $\overline{24}$, $\overline{34}$, $\overline{8}$, $\overline{18}$, $\overline{28}$, $\overline{2}$, $\overline{12}$, $\overline{22}$, $\overline{32}$, $\overline{6}$, $\overline{16}$, $\overline{26}$, $\overline{36} = \overline{0}$ 18
$\overline{12}$ Multiples of $\overline{12}$ are $\overline{12}$, $\overline{24}$, $\overline{36} = \overline{0}$ 3
$\overline{-1}$ 36 is the smallest multiple of $-1$ that is congruent to 0 mod 36. 36
$\overline{-10}$ Multiples of $\overline{-10}$ are $\overline{-10} = \overline{26}$, $\overline{16}$, $\overline{6}$, $\overline{32}$, $\overline{22}$, $\overline{12}$, $\overline{2}$, $\overline{28}$, $\overline{18}$, $\overline{8}$, $\overline{34}$, $\overline{24}$, $\overline{14}$, $\overline{4}$, $\overline{30}$, $\overline{20}$, $\overline{10}$, $\overline{0}$ 18
$\overline{-18}$ $\overline{-18} + \overline{-18} = \overline{0}$ 2

#### Linearity

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