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Question Number 80505 by Rio Michael last updated on 03/Feb/20

$$\mathrm{Given}\mathrm{that}{7}^k\equiv 1\left(\mathrm{mod}15\right)$$$$\mathrm{a})\mathrm{Write}\mathrm{down}\mathrm{three}\mathrm{values}\mathrm{of}k.$$$$\mathrm{b})\mathrm{Find}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{of}$$$$\mathrm{the}\mathrm{equation}{7}^k\equiv 1\left(mod15\right)$$

Commented by mr W last updated on 03/Feb/20

$$k=4nwithn\in \mathbb{N}$$

Commented by Rio Michael last updated on 03/Feb/20

$$\mathrm{how}\mathrm{sir}?$$

Commented by mr W last updated on 03/Feb/20

$$7^k=15m+1$$$$lastdigitof{7}^{k}canbe7,9,3,1$$$$lastdigitof15m+1canbe6,1$$$$theonlypossibilityisthatbothhave$$$$thelastdigit1.thatmeans{7}^k={\left(7^2\times 7^2\right)}^n,$$$$i.e.k=4n.wecanprovethat{7}^{4n}-1$$$$isindeedamultipleof15.$$

Commented by Rio Michael last updated on 03/Feb/20

$$sirhowdoyougetthierlast$$$$digitsplease$$

Commented by mr W last updated on 03/Feb/20

$$7\Rightarrow 7$$$$7\times 7\Rightarrow 9$$$$7\times 7\times 7\Rightarrow 3$$$$7\times 7\times 7\times 7\Rightarrow 1$$$$7\times 7\times 7\times 7\times 7\Rightarrow 7etc.$$$$$$$$15\times 1+1\Rightarrow 6$$$$15\times 2+1\Rightarrow 1$$$$15\times 3+1\Rightarrow 6etc.$$

Commented by mr W last updated on 04/Feb/20

$$heretheproofthat{7}^{4n}\equiv 1mod\left(15\right)$$$$7^{4n}={2401}^n={(2400+1)}^n=\underset{r=0}{\sum ^n}{C}_r^n\times {2400}^r$$$$=1+\underset{r=1}{\sum ^n}{C}_r^n\times {2400}^r$$$$=1+\underset{r=1}{\sum ^n}{C}_r^n\times {\left(160\times 15\right)}^r$$$$\equiv 1mod\left(15\right)$$

Commented by Rio Michael last updated on 04/Feb/20

$$thankssomuchsir$$

Commented by mr W last updated on 04/Feb/20

$$seealsoQ80580$$

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