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Question Number 178282 by Sheshdevsahu last updated on 14/Oct/22

$$$$$$findunitdigitof$$$$1^1+2^2+3^3+.......+{63}^{63}+{64}^{64}$$

Answered by mr W last updated on 14/Oct/22

$${(...0)}^n=...0$$$${10}^{10}+{20}^{20}+{30}^{30}+{40}^{49}+{50}^{50}+{60}^{60}=...0$$$${(...1)}^n=...1$$$$1^1+{11}^{11}+{21}^{21}+{31}^{31}+{41}^{41}+{51}^{51}+{61}^{61}=...1\times 7=...7$$$${(...5)}^n=...5$$$$5^5+{15}^{15}+{25}^{25}+{35}^{35}+{45}^{45}+{55}^{55}=...5\times 6=...0$$$${(...6)}^n=...6$$$$6^6+{16}^{16}+{26}^{26}+{36}^{36}+{46}^{46}+{56}^{56}=...6\times 6=...6$$$${(...2)}^{4n+1/2/3/4}=...2/4/8/6$$$$2^2+{12}^{12}+{22}^{22}+{32}^{32}+{42}^{42}+{52}^{52}+{62}^{62}=...4+6+4+6+4+6+4=...4$$$${(...3)}^{4n+1/2/3/4}=...3/9/7/1$$$$3^3+{13}^{13}+{23}^{23}+{33}^{33}+{43}^{43}+{53}^{53}+{63}^{63}=...7+3+7+3+7+3+7=...7$$$${(...4)}^{2n+1/2}=...4/6$$$$4^4+{14}^{14}+{24}^{24}+{34}^{34}+{44}^{44}+{54}^{54}+{64}^{64}=...6\times 7=...2$$$${(...7)}^{4n+1/2/3/4}=...7/9/3/1$$$$7^7+{17}^{17}+{27}^{27}+{37}^{37}+{47}^{47}+{57}^{57}=...3+7+3+7+3+7=...0$$$${(...8)}^{4n+1/2/3/4}=...8/4/2/6$$$$8^8+{18}^{18}+{28}^{28}+{38}^{38}+{48}^{48}+{58}^{58}=...6+4+6+4+6+4=...0$$$${(...9)}^{2n+1/2}=...9/1$$$$9^9+{19}^{19}+{29}^{29}+{39}^{39}+{49}^{49}+{59}^{59}=...9\times 6=...4$$$$$$$$1^1+2^2+...+{64}^{64}=...0+7+0+6+4+7+2+0+0+4=...0$$

Commented by mr W last updated on 14/Oct/22

$$maybetherearebetterways.$$

Commented by Tawa11 last updated on 14/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

Commented by mr W last updated on 14/Oct/22

Answered by mahdipoor last updated on 14/Oct/22

$$lemmaI:$$$${(20+n)}^{(20+n)}\stackrel{10}{\equiv }{n}^{20+n}\stackrel{10}{\equiv }{n}^n$$$$\Rightarrow A\equiv 1^1+...{64}^{64}\equiv mod10$$$$(1^1+...+{20}^{20})+({21}^{21}+...{40}^{40})+({41}^{41}+...+{60}^{60})$$$$+{61}^{61}+{62}^{62}+{63}^{63}+{64}^{64}$$$$\equiv 3\times (1^1+...+{20}^{20})+1^1+2^2+3^3+4^4$$$$lemmaII:$$$$1^1+....+{20}^{20}\equiv mod10$$$$(1^1+{11}^{11})+(2^2+{12}^{12})+...+({10}^{10}+{20}^{20})\equiv 5$$$$lemmaI+II:$$$$\Rightarrow A\equiv 3\times 5+1+4+7+6\equiv 3mod10$$

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