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evaluate-1-2-3-3-4-5-mod-10-




Question Number 215 by 123456 last updated on 25/Jan/15
evaluate  1+2^3 +3^4^5  (mod 10)
$$\mathrm{evaluate} \\ $$$$\mathrm{1}+\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{4}^{\mathrm{5}} } \left(\mathrm{mod}\:\mathrm{10}\right) \\ $$
Answered by prakash jain last updated on 16/Dec/14
1+2^3 =9  3^4^5   is an even power of 9 (9^(512) ) so  units place of 3^4^5    is 1.  unit place of 1+2^3 +3^4^5   is 0  hence 1+2^3 +3^4^5   (mod 10)=0
$$\mathrm{1}+\mathrm{2}^{\mathrm{3}} =\mathrm{9} \\ $$$$\mathrm{3}^{\mathrm{4}^{\mathrm{5}} } \:\mathrm{is}\:\mathrm{an}\:\mathrm{even}\:\mathrm{power}\:\mathrm{of}\:\mathrm{9}\:\left(\mathrm{9}^{\mathrm{512}} \right)\:\mathrm{so} \\ $$$$\mathrm{units}\:\mathrm{place}\:\mathrm{of}\:\mathrm{3}^{\mathrm{4}^{\mathrm{5}} } \:\:\mathrm{is}\:\mathrm{1}. \\ $$$$\mathrm{unit}\:\mathrm{place}\:\mathrm{of}\:\mathrm{1}+\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{4}^{\mathrm{5}} } \:\mathrm{is}\:\mathrm{0} \\ $$$$\mathrm{hence}\:\mathrm{1}+\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{4}^{\mathrm{5}} } \:\left(\mathrm{mod}\:\mathrm{10}\right)=\mathrm{0} \\ $$