Question and Answers Forum

All Questions      Topic List

Number Theory Questions

Previous in All Question      Next in All Question      

Previous in Number Theory      Next in Number Theory      

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

find unit digit of 1^1 +2^2 +3^3 +.......+63^(63) +64^(64)

$$ \\ $$$${find}\:{unit}\:{digit}\:{of}\: \\ $$$$\mathrm{1}^{\mathrm{1}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{3}} +.......+\mathrm{63}^{\mathrm{63}} +\mathrm{64}^{\mathrm{64}} \\ $$

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) =...1×7=...7 (...5)^n =...5 5^5 +15^(15) +25^(25) +35^(35) +45^(45) +55^(55) =...5×6=...0 (...6)^n =...6 6^6 +16^(16) +26^(26) +36^(36) +46^(46) +56^(56) =...6×6=...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) =...6×7=...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) =...9×6=...4 1^1 +2^2 +...+64^(64) =...0+7+0+6+4+7+2+0+0+4=...0

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

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

maybe there are better ways.

$${maybe}\:{there}\:{are}\:{better}\:{ways}. \\ $$

Commented by Tawa11 last updated on 14/Oct/22

Great sir

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

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

Answered by mahdipoor last updated on 14/Oct/22

lemma I: (20+n)^((20+n)) ≡^(10) n^(20+n) ≡^(10) n^n ⇒A≡1^1 +...64^(64) ≡ mod 10 (1^1 +...+20^(20) )+(21^(21) +...40^(40) )+(41^(41) +...+60^(60) ) +61^(61) +62^(62) +63^(63) +64^(64) ≡3×(1^1 +...+20^(20) )+1^1 +2^2 +3^3 +4^4 lemma II: 1^1 +....+20^(20) ≡ mod 10 (1^1 +11^(11) )+(2^2 +12^(12) )+...+(10^(10) +20^(20) )≡5 lemma I + II : ⇒A≡3×5+1+4+7+6≡3 mod 10

$${lemma}\:{I}: \\ $$$$\left(\mathrm{20}+{n}\right)^{\left(\mathrm{20}+{n}\right)} \overset{\mathrm{10}} {\equiv}{n}^{\mathrm{20}+{n}} \overset{\mathrm{10}} {\equiv}{n}^{{n}} \:\:\: \\ $$$$\Rightarrow{A}\equiv\mathrm{1}^{\mathrm{1}} +...\mathrm{64}^{\mathrm{64}} \equiv\:\:\:\:\:\:\:\:{mod}\:\mathrm{10} \\ $$$$\left(\mathrm{1}^{\mathrm{1}} +...+\mathrm{20}^{\mathrm{20}} \right)+\left(\mathrm{21}^{\mathrm{21}} +...\mathrm{40}^{\mathrm{40}} \right)+\left(\mathrm{41}^{\mathrm{41}} +...+\mathrm{60}^{\mathrm{60}} \right) \\ $$$$+\mathrm{61}^{\mathrm{61}} +\mathrm{62}^{\mathrm{62}} +\mathrm{63}^{\mathrm{63}} +\mathrm{64}^{\mathrm{64}} \\ $$$$\equiv\mathrm{3}×\left(\mathrm{1}^{\mathrm{1}} +...+\mathrm{20}^{\mathrm{20}} \right)+\mathrm{1}^{\mathrm{1}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{3}} +\mathrm{4}^{\mathrm{4}} \:\:\: \\ $$$${lemma}\:{II}: \\ $$$$\mathrm{1}^{\mathrm{1}} +....+\mathrm{20}^{\mathrm{20}} \equiv\:\:\:\:\:\:{mod}\:\mathrm{10} \\ $$$$\left(\mathrm{1}^{\mathrm{1}} +\mathrm{11}^{\mathrm{11}} \right)+\left(\mathrm{2}^{\mathrm{2}} +\mathrm{12}^{\mathrm{12}} \right)+...+\left(\mathrm{10}^{\mathrm{10}} +\mathrm{20}^{\mathrm{20}} \right)\equiv\mathrm{5} \\ $$$${lemma}\:{I}\:+\:{II}\:: \\ $$$$\Rightarrow{A}\equiv\mathrm{3}×\mathrm{5}+\mathrm{1}+\mathrm{4}+\mathrm{7}+\mathrm{6}\equiv\mathrm{3}\:\:\:\:\:{mod}\:\mathrm{10} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com