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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

$$lastdigitis5.$$

Answered by ArielVyny last updated on 01/Jul/21

$$U_k=\underset{k=1}{\sum ^{1989}}{k}^{1989}$$$$ln\left(U_k\right)=\underset{k=1}{\sum ^{1989}}1989ln\left(k\right)=1989\underset{k=1}{\sum ^{1989}}ln\left(k\right)=1989ln(k!)$$$$U_k=e^{1989ln(k!)}={(k!)}^{1989}$$$$$$

Commented by mathdanisur last updated on 01/Jul/21

$$thankyouSer,thentheanswer?$$

Commented by mr W last updated on 01/Jul/21

$$\ln (a^n+b^n)\ne n\ln a+n\ln b$$

Commented by mr W last updated on 02/Jul/21

$$1^{1989}+2^{1989}+3^{1989}+...+{1989}^{1989}$$$$\ne {(1\times 2\times 3\times ...\times 1989)}^{1989}$$

Answered by mr W last updated on 01/Jul/21

$$1:\Rightarrow 1$$$$2:2/4/8/6/2\Rightarrow 2$$$$3:3/9/7/1/3\Rightarrow 3$$$$4:4/6/4\Rightarrow 4$$$$5:\Rightarrow 5$$$$6:\Rightarrow 6$$$$7:7/9/3/1/7\Rightarrow 7$$$$8:8/4/2/6/8\Rightarrow 8$$$$9:9/1/9\Rightarrow 9$$$$0:\Rightarrow 0$$$$1:1,11,21,...,1971,1981\Rightarrow 199\times 1=9$$$$2:2,12,22,...,1972,1982\Rightarrow 199\times 2=8$$$$3:\Rightarrow 199\times 3=7$$$$4:\Rightarrow 199\times 4=6$$$$5:\Rightarrow 199\times 5=5$$$$6:\Rightarrow 199\times 6=4$$$$7:\Rightarrow 199\times 7=3$$$$8:\Rightarrow 199\times 8=2$$$$9:\Rightarrow 199\times 9=1$$$$1+2+3+...+9=5$$

Commented by mathdanisur last updated on 01/Jul/21

$$alotcoolSer,thankyou$$

Answered by Rasheed.Sindhi last updated on 02/Jul/21

$$(1^{1989}+2^{1989}+3^{1989}+...+{1989}^{1989})\left(\mathrm{mod}10\right)$$$$\mathcal{B}yobservingvariouspowersof$$$$variousnumbersweconclude:$$$${\mathrm{a}}^{4\mathrm{n}+1}\equiv \mathrm{a}\left(\mathrm{mod}10\right)$$$${\mathrm{a}}^{1989}={\mathrm{a}}^{497\times 4+1}\equiv \mathrm{a}\left(\mathrm{mod}10\right)$$$$1^{1989}+2^{1989}+3^{1989}+...+{1989}^{1989}\left(\mathrm{mod}10\right)$$$$=1^{497\times 4+1}+2^{497\times 4+1}+...{1989}^{497\times 4+1}(\mathrm{mod}10$$$$=1+2+3+...+1989\left(\mathrm{mod}10\right)$$$$=\frac{1989\times 1990}{2}\left(\mathrm{mod}10\right)$$$$=1979055\left(\mathrm{mod}10\right)$$$$=5$$

Commented by mathdanisur last updated on 02/Jul/21

$$ThankyouSer,alotcool$$

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