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Question-145047

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Question Number 145047 by mathdanisur last updated on 01/Jul/21
Commented by mr W last updated on 01/Jul/21
last digit is 5.
$${last}\:{digit}\:{is}\:\mathrm{5}. \\ $$
Answered by ArielVyny last updated on 01/Jul/21
U_k =Σ_(k=1) ^(1989) k^(1989) ln(U_k )=Σ_(k=1) ^(1989) 1989ln(k)=1989Σ_(k=1) ^(1989) ln(k)=1989ln(k!) U_k =e^(1989ln(k!)) =(k!)^(1989)
$${U}_{{k}} =\underset{{k}=\mathrm{1}} {\overset{\mathrm{1989}} {\sum}}{k}^{\mathrm{1989}} \\ $$$${ln}\left({U}_{{k}} \right)=\underset{{k}=\mathrm{1}} {\overset{\mathrm{1989}} {\sum}}\mathrm{1989}{ln}\left({k}\right)=\mathrm{1989}\underset{{k}=\mathrm{1}} {\overset{\mathrm{1989}} {\sum}}{ln}\left({k}\right)=\mathrm{1989}{ln}\left({k}!\right) \\ $$$${U}_{{k}} ={e}^{\mathrm{1989}{ln}\left({k}!\right)} =\left({k}!\right)^{\mathrm{1989}} \\ $$$$ \\ $$
Commented by mathdanisur last updated on 01/Jul/21
thank you Ser, then the answer?
$${thank}\:{you}\:{Ser},\:{then}\:{the}\:{answer}? \\ $$
Commented by mr W last updated on 01/Jul/21
ln (a^n +b^n )≠nln a+nln b
$$\mathrm{ln}\:\left({a}^{{n}} +{b}^{{n}} \right)\neq{n}\mathrm{ln}\:{a}+{n}\mathrm{ln}\:{b} \\ $$
Commented by mr W last updated on 02/Jul/21
1^(1989) +2^(1989) +3^(1989) +...+1989^(1989) ≠(1×2×3×...×1989)^(1989)
$$\mathrm{1}^{\mathrm{1989}} +\mathrm{2}^{\mathrm{1989}} +\mathrm{3}^{\mathrm{1989}} +…+\mathrm{1989}^{\mathrm{1989}} \\ $$$$\neq\left(\mathrm{1}×\mathrm{2}×\mathrm{3}×…×\mathrm{1989}\right)^{\mathrm{1989}} \\ $$
Answered by mr W last updated on 01/Jul/21
1: ⇒ 1 2: 2/4/8/6/2 ⇒2 3: 3/9/7/1/3 ⇒ 3 4: 4/6/4 ⇒4 5: ⇒5 6: ⇒ 6 7: 7/9/3/1/7 ⇒7 8: 8/4/2/6/8 ⇒ 8 9: 9/1/9 ⇒9 0: ⇒ 0 1: 1,11,21,...,1971,1981 ⇒199×1=9 2: 2,12,22,...,1972,1982 ⇒199×2=8 3: ⇒199×3=7 4: ⇒199×4=6 5: ⇒ 199×5=5 6: ⇒ 199×6=4 7: ⇒ 199×7=3 8: ⇒ 199×8=2 9: ⇒ 199×9=1 1+2+3+...+9=5
$$\mathrm{1}:\:\Rightarrow\:\mathrm{1} \\ $$$$\mathrm{2}:\:\mathrm{2}/\mathrm{4}/\mathrm{8}/\mathrm{6}/\mathrm{2}\:\Rightarrow\mathrm{2} \\ $$$$\mathrm{3}:\:\mathrm{3}/\mathrm{9}/\mathrm{7}/\mathrm{1}/\mathrm{3}\:\Rightarrow\:\mathrm{3} \\ $$$$\mathrm{4}:\:\mathrm{4}/\mathrm{6}/\mathrm{4}\:\Rightarrow\mathrm{4} \\ $$$$\mathrm{5}:\:\Rightarrow\mathrm{5} \\ $$$$\mathrm{6}:\:\Rightarrow\:\mathrm{6} \\ $$$$\mathrm{7}:\:\mathrm{7}/\mathrm{9}/\mathrm{3}/\mathrm{1}/\mathrm{7}\:\Rightarrow\mathrm{7} \\ $$$$\mathrm{8}:\:\mathrm{8}/\mathrm{4}/\mathrm{2}/\mathrm{6}/\mathrm{8}\:\Rightarrow\:\mathrm{8} \\ $$$$\mathrm{9}:\:\mathrm{9}/\mathrm{1}/\mathrm{9}\:\Rightarrow\mathrm{9} \\ $$$$\mathrm{0}:\:\Rightarrow\:\mathrm{0} \\ $$$$\mathrm{1}:\:\mathrm{1},\mathrm{11},\mathrm{21},…,\mathrm{1971},\mathrm{1981}\:\Rightarrow\mathrm{199}×\mathrm{1}=\mathrm{9} \\ $$$$\mathrm{2}:\:\mathrm{2},\mathrm{12},\mathrm{22},…,\mathrm{1972},\mathrm{1982}\:\Rightarrow\mathrm{199}×\mathrm{2}=\mathrm{8} \\ $$$$\mathrm{3}:\:\Rightarrow\mathrm{199}×\mathrm{3}=\mathrm{7} \\ $$$$\mathrm{4}:\:\Rightarrow\mathrm{199}×\mathrm{4}=\mathrm{6} \\ $$$$\mathrm{5}:\:\Rightarrow\:\mathrm{199}×\mathrm{5}=\mathrm{5} \\ $$$$\mathrm{6}:\:\Rightarrow\:\mathrm{199}×\mathrm{6}=\mathrm{4} \\ $$$$\mathrm{7}:\:\Rightarrow\:\mathrm{199}×\mathrm{7}=\mathrm{3} \\ $$$$\mathrm{8}:\:\Rightarrow\:\mathrm{199}×\mathrm{8}=\mathrm{2} \\ $$$$\mathrm{9}:\:\Rightarrow\:\mathrm{199}×\mathrm{9}=\mathrm{1} \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}+…+\mathrm{9}=\mathrm{5} \\ $$
Commented by mathdanisur last updated on 01/Jul/21
a lot cool Ser, thank you
$${a}\:{lot}\:{cool}\:{Ser},\:{thank}\:{you} \\ $$
Answered by Rasheed.Sindhi last updated on 02/Jul/21
(1^(1989) +2^(1989) +3^(1989) +...+1989^(1989) )(mod10) By observing various powers of various numbers we conclude: a^(4n+1) ≡a(mod 10) a^(1989) =a^(497×4+1) ≡a(mod 10) 1^(1989) +2^(1989) +3^(1989) +...+1989^(1989) (mod10) =1^(497×4+1) +2^(497×4+1) +...1989^(497×4+1) (mod 10 =1+2+3+...+1989(mod 10) =((1989×1990)/2)(mod 10) =1979055(mod 10) =5
$$\left(\mathrm{1}^{\mathrm{1989}} +\mathrm{2}^{\mathrm{1989}} +\mathrm{3}^{\mathrm{1989}} +…+\mathrm{1989}^{\mathrm{1989}} \right)\left(\mathrm{mod10}\right) \\ $$$$\mathcal{B}{y}\:{observing}\:{various}\:{powers}\:{of}\: \\ $$$${various}\:{numbers}\:{we}\:{conclude}: \\ $$$$\mathrm{a}^{\mathrm{4n}+\mathrm{1}} \equiv\mathrm{a}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\mathrm{a}^{\mathrm{1989}} =\mathrm{a}^{\mathrm{497}×\mathrm{4}+\mathrm{1}} \equiv\mathrm{a}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\mathrm{1}^{\mathrm{1989}} +\mathrm{2}^{\mathrm{1989}} +\mathrm{3}^{\mathrm{1989}} +…+\mathrm{1989}^{\mathrm{1989}} \left(\mathrm{mod10}\right) \\ $$$$=\mathrm{1}^{\mathrm{497}×\mathrm{4}+\mathrm{1}} +\mathrm{2}^{\mathrm{497}×\mathrm{4}+\mathrm{1}} +…\mathrm{1989}^{\mathrm{497}×\mathrm{4}+\mathrm{1}} \left(\mathrm{mod}\:\mathrm{10}\right. \\ $$$$=\mathrm{1}+\mathrm{2}+\mathrm{3}+…+\mathrm{1989}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$=\frac{\mathrm{1989}×\mathrm{1990}}{\mathrm{2}}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$=\mathrm{1979055}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$=\mathrm{5} \\ $$
Commented by mathdanisur last updated on 02/Jul/21
Thankyou Ser, a lot cool
$${Thankyou}\:{Ser},\:{a}\:{lot}\:{cool} \\ $$

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