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Question Number 137364 by liberty last updated on 02/Apr/21

$$Findtheremainder{7}^{30}divide$$$$by10$$

Answered by MJS_new last updated on 02/Apr/21

$$7^0=1$$$$7^1=7$$$$7^2=49$$$$7^3=343$$$$7^4=2401$$$$30=7\times 4+2\Rightarrow \mathrm{answer}\mathrm{is}9$$

Answered by EDWIN88 last updated on 02/Apr/21

$$\mathrm{we}\mathrm{have}{7}^{30}={49}^{15}\equiv 9^{15}\left(\mathrm{mod}10\right)$$$$\mathrm{we}\mathrm{can}\mathrm{apply}\mathrm{Binomial}\mathrm{Theorem}$$$$9^{15}={(10-1)}^{15}=\underset{\mathrm{k}=0}{\sum ^{15}}\left(\begin{array}{c}15\\ \mathrm{k}\end{array}\right){10}^{15-\mathrm{k}}.{(-1)}^{\mathrm{k}}$$$$\mathrm{so}\mathrm{clearly}{9}^{15}\equiv -1\left(\mathrm{mod}10\right)=9\left(\mathrm{mod}10\right)$$$$\mathrm{therefore}\mathrm{the}\mathrm{remainder}{7}^{30}:10\mathrm{is}\mathrm{equal}$$$$\mathrm{to}9.$$

Answered by benjo_mathlover last updated on 02/Apr/21

$$ByEulerphiTheorem$$$$7^{\phi \left(n\right)}=1\left(modn\right)$$$$wefind\phi \left(10\right)=10(1-\frac{1}{2})(1-\frac{1}{5})=4$$$$so{7}^{30}={\left(7^4\right)}^7\times 7^2=1\times 7^2\left(mod10\right)$$$$=49\left(mod10\right)=9\left(mod10\right)$$$$itfollowsthattheremainder$$$$is9$$

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