Question Number 191862 by cortano12 last updated on 02/May/23
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{last}\:\mathrm{digit}\:\mathrm{from}\: \\ $$$$\:\left(\mathrm{2}^{\mathrm{400}} −\mathrm{2}^{\mathrm{320}} \right)\left(\mathrm{2}^{\mathrm{200}} +\mathrm{2}^{\mathrm{160}} \right)\left(\mathrm{2}^{\mathrm{200}} −\mathrm{2}^{\mathrm{160}} \right) \\ $$
Commented by BaliramKumar last updated on 02/May/23
$$\mathrm{0} \\ $$
Answered by Subhi last updated on 02/May/23
$$ \\ $$$$\left(\mathrm{2}^{\mathrm{400}} −\mathrm{2}^{\mathrm{320}} \right)^{\mathrm{2}} \\ $$$$\mathrm{2}^{\mathrm{400}} {mod}\mathrm{10}\: \\ $$$$\mathrm{2}^{\mathrm{4}} \:\:\Xi\mathrm{6}{mod}\mathrm{10} \\ $$$$\mathrm{2}^{\mathrm{400}} \Xi\mathrm{6}^{\mathrm{100}} {mod}\mathrm{10}\Xi\mathrm{6}^{\mathrm{5}} {mod}\mathrm{10}\Xi\mathrm{6}{mod}\mathrm{10} \\ $$$$\mathrm{6}^{\mathrm{5}} =\mathrm{6}^{\mathrm{4}} ×\mathrm{6}\Xi\mathrm{6}^{\mathrm{2}} {mod}\mathrm{10}\Xi\mathrm{6}{mod}\mathrm{10} \\ $$$${by}\:{the}\:{same}\:{way}\: \\ $$$$\mathrm{3}^{\mathrm{320}} \Xi\mathrm{6}^{\mathrm{80}} \Xi\mathrm{6}{mod}\mathrm{10} \\ $$$${thd}\:{last}\:{digit}=\:\left(\mathrm{6}−\mathrm{6}\right)^{\mathrm{2}} =\mathrm{0} \\ $$