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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