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Question Number 795 by malwaan last updated on 15/Mar/15

$$whatisthelastdigitof$$$$7^{\left(7^{(7....)}\right)}$$$$thenumberof{7}^{\prime }sis1001$$

Answered by prakash jain last updated on 15/Mar/15

$$7^1\equiv 3\left(\mathrm{mod}4\right),7^2\equiv 1\left(\mathrm{mod}4\right)$$$$7^7\equiv 3^4{3}^3\equiv 27\equiv 3\left(\mathrm{mod}4\right)$$$$7^7\equiv 3^{4k+3}\equiv {\left(3^4\right)}^k{3}^3\equiv 3\left(\mathrm{mod}4\right)$$$$7^{7^{7\left(1000\mathrm{times}\right)}}\equiv 3\left(\mathrm{mod}4\right)$$$$7^{4\mathrm{k}+3}=7^{4k}{7}^3$$$$7^{4k}\mathrm{has}\mathrm{last}\mathrm{digit}1.$$$$7^3\mathrm{has}\mathrm{last}\mathrm{digits}3.$$$$7^{4k+3}\mathrm{will}\mathrm{have}\mathrm{last}\mathrm{digit}3.$$

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