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Question Number 43072 by Raj Singh last updated on 07/Sep/18

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Sep/18

3^(50) (3^(49) −1)  3^1 →unit term is 3  3^2 →unit term is9  3^3 →unit term is 7  3^4 →unit term is1  3^5 →unit term is 3    so sequence is 3 9 7 1   3 9 7 1  now  3^(50) (3^(49) −1)  (3^4 )^(12) ×3^2 {(3^4 )^(12) ×3−1}  unit term for 3^(48) →(3^4 )^(12)  is1  unit term of 3^(48) ×3^2 →9  unit term of (3^4 )^(12) ×3→3  unit term of (3^4 )^(12) ×3−1 →2  so unit term of  (3^4 )^(12) ×3^2 {(3^4 )^(12) ×3−1} is→8

$$\mathrm{3}^{\mathrm{50}} \left(\mathrm{3}^{\mathrm{49}} −\mathrm{1}\right) \\ $$$$\mathrm{3}^{\mathrm{1}} \rightarrow{unit}\:{term}\:{is}\:\mathrm{3} \\ $$$$\mathrm{3}^{\mathrm{2}} \rightarrow{unit}\:{term}\:{is}\mathrm{9} \\ $$$$\mathrm{3}^{\mathrm{3}} \rightarrow{unit}\:{term}\:{is}\:\mathrm{7} \\ $$$$\mathrm{3}^{\mathrm{4}} \rightarrow{unit}\:{term}\:{is}\mathrm{1} \\ $$$$\mathrm{3}^{\mathrm{5}} \rightarrow{unit}\:{term}\:{is}\:\mathrm{3} \\ $$$$ \\ $$$${so}\:{sequence}\:{is}\:\mathrm{3}\:\mathrm{9}\:\mathrm{7}\:\mathrm{1}\:\:\:\mathrm{3}\:\mathrm{9}\:\mathrm{7}\:\mathrm{1} \\ $$$${now} \\ $$$$\mathrm{3}^{\mathrm{50}} \left(\mathrm{3}^{\mathrm{49}} −\mathrm{1}\right) \\ $$$$\left(\mathrm{3}^{\mathrm{4}} \right)^{\mathrm{12}} ×\mathrm{3}^{\mathrm{2}} \left\{\left(\mathrm{3}^{\mathrm{4}} \right)^{\mathrm{12}} ×\mathrm{3}−\mathrm{1}\right\} \\ $$$${unit}\:{term}\:{for}\:\mathrm{3}^{\mathrm{48}} \rightarrow\left(\mathrm{3}^{\mathrm{4}} \right)^{\mathrm{12}} \:{is}\mathrm{1} \\ $$$${unit}\:{term}\:{of}\:\mathrm{3}^{\mathrm{48}} ×\mathrm{3}^{\mathrm{2}} \rightarrow\mathrm{9} \\ $$$${unit}\:{term}\:{of}\:\left(\mathrm{3}^{\mathrm{4}} \right)^{\mathrm{12}} ×\mathrm{3}\rightarrow\mathrm{3} \\ $$$${unit}\:{term}\:{of}\:\left(\mathrm{3}^{\mathrm{4}} \right)^{\mathrm{12}} ×\mathrm{3}−\mathrm{1}\:\rightarrow\mathrm{2} \\ $$$${so}\:{unit}\:{term}\:{of} \\ $$$$\left(\mathrm{3}^{\mathrm{4}} \right)^{\mathrm{12}} ×\mathrm{3}^{\mathrm{2}} \left\{\left(\mathrm{3}^{\mathrm{4}} \right)^{\mathrm{12}} ×\mathrm{3}−\mathrm{1}\right\}\:{is}\rightarrow\mathrm{8} \\ $$

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