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Question-54271




Question Number 54271 by naveen200601 last updated on 01/Feb/19
Commented by Meritguide1234 last updated on 02/Feb/19
post many problem on this type
$$\mathrm{post}\:\mathrm{many}\:\mathrm{problem}\:\mathrm{on}\:\mathrm{this}\:\mathrm{type} \\ $$
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  last digit (3,9,7,1) is repeating in cycle wise..  3^(4n) =(81)^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 ×3^(nc_1 80) ×3^(nc_2 80^2 ) ...×3^(80^n )   =(unit place 3)×(unit place 1)(u.p 1)...(u.p 1)  =so unit place is 3  so u.p of 3^3^(4n)  +1=3+1=4
$$\mathrm{3}^{\mathrm{4}} =\mathrm{81} \\ $$$$\mathrm{3}^{\mathrm{1}} =\mathrm{3} \\ $$$$\mathrm{3}^{\mathrm{2}} =\mathrm{9} \\ $$$$\mathrm{3}^{\mathrm{3}} =\mathrm{27} \\ $$$$\mathrm{3}^{\mathrm{4}} =\mathrm{81} \\ $$$$\mathrm{3}^{\mathrm{5}} =\mathrm{243} \\ $$$$\mathrm{3}^{\mathrm{6}} =\mathrm{729} \\ $$$$…{thus} \\ $$$${last}\:{digit}\:\left(\mathrm{3},\mathrm{9},\mathrm{7},\mathrm{1}\right)\:{is}\:{repeating}\:{in}\:{cycle}\:{wise}.. \\ $$$$\mathrm{3}^{\mathrm{4}{n}} =\left(\mathrm{81}\right)^{{n}} =\left(\mathrm{1}+\mathrm{80}\right)^{{n}} \\ $$$$\mathrm{1}+{nc}_{\mathrm{1}} \mathrm{80}+{nc}_{\mathrm{2}} \mathrm{80}^{\mathrm{2}} +{nc}_{\mathrm{3}} \mathrm{80}^{\mathrm{3}} +…+\mathrm{80}^{{n}} \\ $$$$\mathrm{3}^{\mathrm{3}^{\mathrm{4}{n}} } \\ $$$$=\mathrm{3}^{\left(\mathrm{1}+{nc}_{\mathrm{1}} \mathrm{80}+{nc}_{\mathrm{2}} \mathrm{80}^{\mathrm{2}} +{nc}_{\mathrm{3}} \mathrm{80}^{\mathrm{3}} +…+\mathrm{80}^{{n}} \right.} \\ $$$$=\mathrm{3}^{\mathrm{1}} ×\mathrm{3}^{{nc}_{\mathrm{1}} \mathrm{80}} ×\mathrm{3}^{{nc}_{\mathrm{2}} \mathrm{80}^{\mathrm{2}} } …×\mathrm{3}^{\mathrm{80}^{{n}} } \\ $$$$=\left({unit}\:{place}\:\mathrm{3}\right)×\left({unit}\:{place}\:\mathrm{1}\right)\left({u}.{p}\:\mathrm{1}\right)…\left({u}.{p}\:\mathrm{1}\right) \\ $$$$={so}\:{unit}\:{place}\:{is}\:\mathrm{3} \\ $$$${so}\:{u}.{p}\:{of}\:\mathrm{3}^{\mathrm{3}^{\mathrm{4}{n}} } +\mathrm{1}=\mathrm{3}+\mathrm{1}=\mathrm{4} \\ $$

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