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Question Number 191862 by cortano12 last updated on 02/May/23

$$\mathrm{Find}\mathrm{the}\mathrm{last}\mathrm{digit}\mathrm{from}$$$$(2^{400}-2^{320})(2^{200}+2^{160})(2^{200}-2^{160})$$

Commented by BaliramKumar last updated on 02/May/23

$$0$$

Answered by Subhi last updated on 02/May/23

$$$$$${(2^{400}-2^{320})}^2$$$$2^{400}mod10$$$$2^4\mathrm{\Xi }6mod10$$$$2^{400}\mathrm{\Xi }{6}^{100}mod10\mathrm{\Xi }{6}^{5}mod10\mathrm{\Xi }6mod10$$$$6^5=6^4\times 6\mathrm{\Xi }{6}^{2}mod10\mathrm{\Xi }6mod10$$$$bythesameway$$$$3^{320}\mathrm{\Xi }{6}^{80}\mathrm{\Xi }6mod10$$$$thdlastdigit={(6-6)}^2=0$$

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