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

what is the last digit of  7^((7^((7....)) ))    the number of 7′s is 1001

$${what}\:{is}\:{the}\:{last}\:{digit}\:{of} \\ $$$$\mathrm{7}^{\left(\mathrm{7}^{\left(\mathrm{7}....\right)} \right)} \: \\ $$$${the}\:{number}\:{of}\:\mathrm{7}'{s}\:{is}\:\mathrm{1001} \\ $$

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

7^1 ≡3 (mod 4), 7^2 ≡1 (mod 4)  7^7 ≡3^4 3^3 ≡27≡3 (mod 4)  7^7 ≡3^(4k+3) ≡(3^4 )^k 3^3  ≡ 3 (mod 4)  7^7^(7(1000 times))  ≡3(mod 4)  7^(4k+3) =7^(4k) 7^3   7^(4k)  has last digit 1.  7^3  has last digits 3.  7^(4k+3)  will have last digit 3.

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

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