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Question Number 54271 by naveen200601 last updated on 01/Feb/19

Commented by Meritguide1234 last updated on 02/Feb/19

$$\mathrm{post}\mathrm{many}\mathrm{problem}\mathrm{on}\mathrm{this}\mathrm{type}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Feb/19

$$3^4=81$$$$3^1=3$$$$3^2=9$$$$3^3=27$$$$3^4=81$$$$3^5=243$$$$3^6=729$$$$...thus$$$$lastdigit(3,9,7,1)isrepeatingincyclewise..$$$$3^{4n}={\left(81\right)}^n={(1+80)}^n$$$$1+{nc}_{1}80+{nc}_2{80}^2+{nc}_3{80}^3+...+{80}^n$$$$3^{3^{4n}}$$$$=3^{(1+{nc}_{1}80+{nc}_2{80}^2+{nc}_3{80}^3+...+{80}^n}$$$$=3^1\times 3^{{nc}_{1}80}\times 3^{{nc}_2{80}^2}...\times 3^{{80}^n}$$$$=\left(unitplace3\right)\times \left(unitplace1\right)(u.p1)...(u.p1)$$$$=sounitplaceis3$$$$sou.pof{3}^{3^{4n}}+1=3+1=4$$

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