Question Number 130727 by EDWIN88 last updated on 28/Jan/21
Commented by MJS_new last updated on 28/Jan/21
$$\mathrm{6} \\ $$
Commented by EDWIN88 last updated on 28/Jan/21
$${yes} \\ $$
Answered by liberty last updated on 28/Jan/21
![2^(5k) = 2 (mod 10) ∀k∈Z^+ 320 = 25×12+5×4 2^(320) = [(2^5 )^5 ]^(12) ×(2^5 )^4 (mod 10) = [(2)^5 ]^2 ×(2^4 )×2^2 (mod 10) = 2^2 ×2^6 = 2^5 ×2^3 = 16 (mod 10) = 6 (mod 10)](Q130730.png)
$$\:\mathrm{2}^{\mathrm{5k}} \:=\:\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{10}\right)\:\forall\mathrm{k}\in\mathbb{Z}^{+} \\ $$$$\:\mathrm{320}\:=\:\mathrm{25}×\mathrm{12}+\mathrm{5}×\mathrm{4} \\ $$$$\mathrm{2}^{\mathrm{320}} \:=\:\left[\left(\mathrm{2}^{\mathrm{5}} \right)^{\mathrm{5}} \right]^{\mathrm{12}} ×\left(\mathrm{2}^{\mathrm{5}} \right)^{\mathrm{4}} \left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\left[\left(\mathrm{2}\right)^{\mathrm{5}} \right]^{\mathrm{2}} ×\left(\mathrm{2}^{\mathrm{4}} \right)×\mathrm{2}^{\mathrm{2}} \:\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}^{\mathrm{2}} ×\mathrm{2}^{\mathrm{6}} \:=\:\mathrm{2}^{\mathrm{5}} ×\mathrm{2}^{\mathrm{3}} \:=\:\mathrm{16}\:\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\mathrm{6}\:\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$
Commented by JDamian last updated on 29/Jan/21
$$\mathrm{2}^{\mathrm{5}{k}} \neq\:\mathrm{2}\left({mod}\:\mathrm{10}\right) \\ $$$${k}=\mathrm{2}\:\:\Rightarrow\:\:\mathrm{2}^{\mathrm{10}} =\mathrm{1024} \\ $$
Answered by mr W last updated on 28/Jan/21
$$\mathrm{2}^{{n}} =......\mathrm{2}\:{if}\:{mod}\left({n},\mathrm{4}\right)=\mathrm{1} \\ $$$$\mathrm{2}^{{n}} =......\mathrm{4}\:{if}\:{mod}\left({n},\mathrm{4}\right)=\mathrm{2} \\ $$$$\mathrm{2}^{{n}} =......\mathrm{8}\:{if}\:{mod}\left({n},\mathrm{4}\right)=\mathrm{3} \\ $$$$\mathrm{2}^{{n}} =......\mathrm{6}\:{if}\:{mod}\left({n},\mathrm{4}\right)=\mathrm{0} \\ $$$${since}\:{mod}\left(\mathrm{320},\mathrm{4}\right)=\mathrm{0},\:\mathrm{2}^{\mathrm{320}} \:{ends}\:{with}\:\mathrm{6} \\ $$