Question Number 215 by 123456 last updated on 25/Jan/15
![evaluate 1+2^3 +3^4^5 (mod 10)](https://www.tinkutara.com/question/Q215.png)
$$\mathrm{evaluate} \\ $$$$\mathrm{1}+\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{4}^{\mathrm{5}} } \left(\mathrm{mod}\:\mathrm{10}\right) \\ $$
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](https://www.tinkutara.com/question/Q220.png)
$$\mathrm{1}+\mathrm{2}^{\mathrm{3}} =\mathrm{9} \\ $$$$\mathrm{3}^{\mathrm{4}^{\mathrm{5}} } \:\mathrm{is}\:\mathrm{an}\:\mathrm{even}\:\mathrm{power}\:\mathrm{of}\:\mathrm{9}\:\left(\mathrm{9}^{\mathrm{512}} \right)\:\mathrm{so} \\ $$$$\mathrm{units}\:\mathrm{place}\:\mathrm{of}\:\mathrm{3}^{\mathrm{4}^{\mathrm{5}} } \:\:\mathrm{is}\:\mathrm{1}. \\ $$$$\mathrm{unit}\:\mathrm{place}\:\mathrm{of}\:\mathrm{1}+\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{4}^{\mathrm{5}} } \:\mathrm{is}\:\mathrm{0} \\ $$$$\mathrm{hence}\:\mathrm{1}+\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{4}^{\mathrm{5}} } \:\left(\mathrm{mod}\:\mathrm{10}\right)=\mathrm{0} \\ $$