Question

18 supra 3n+1 = in multimea Z
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Answer 1

$\frac{18}{3n + 1} \: \in \: \mathbb{Z} = > (3n + 1) \: \in \: D_{18}$

$D_{18}=\left \{ \pm1,\pm2,\pm3,\pm6,\pm9,\pm18 \right \}$

$1)3n + 1 = - 1$

$3n = - 1 - 1$

$3n = - 2$

$n = - \frac{2}{3}$

$2)3n + 1 = 1$

$3n = 1 - 1$

$3n = 0 \: | \div 3$

$n = 0$

$3)3n + 1 = - 2$

$3n = - 2 - 1$

$3n = - 3 \: | \div 3$

$n = - 1$

$4)3n + 1 = 2$

$3n = 2 - 1$

$3n = 1$

$n = \frac{1}{3}$

$5)3n + 1 = - 3$

$3n = - 3 - 1$

$3n = - 4$

$n = - \frac{4}{3}$

$6)3n + 1 = 3$

$3n = 3 - 1$

$3n = 2$

$n = \frac{2}{3}$

$7)3n + 1 = - 6$

$3n = - 6 - 1$

$3n = - 7$

$n = - \frac{7}{3}$

$8)3n + 1 = 6$

$3n = 6 - 1$

$3n = 5$

$n = \frac{5}{3}$

$9)3n + 1 = - 9$

$3n = - 9 - 1$

$3n = - 10$

$n = - \frac{10}{3}$

$10)3n + 1 = 9$

$3n = 9 - 1$

$3n = 8$

$n = \frac{8}{3}$

$11)3n + 1 = - 18$

$3n = - 18 - 1$

$3n = - 19$

$n = - \frac{19}{3}$

$12)3n + 1 = 18$

$3n = 18 - 1$

$3n = 17$

$n = \frac{17}{3}$

$n \: \in \: \left \{- \frac{2}{3},0,-1,\frac{1}{3} ,-\frac{4}{3},\frac{2}{3},-\frac{7}{3},\frac{5}{3},-\frac{10}{3},\frac{8}{3},-\frac{19}{3},\frac{17}{3}\right \}$