maths

To solve this problem, find the pattern of the powers of 7. 7^1 is 7, 7^2 is 9(units' digit), and so on.

We'll get...

7

9

3

1

7

9

3

1

and so on...

So 7^2018's unit digit is 9.

~~Hypotenuisance

What is unit digit of 7^2018

\(\begin{array}{|rcll|} \hline && \mathbf{{\color{red}7}^{2018} \pmod{10}}\\ &&&\boxed{ {\color{red}7} \pmod{10} \\ \equiv 7-10 \pmod{10} \\ \equiv {\color{green}-3} \pmod{10} } \\ &\equiv& \left({\color{green}-3}\right)^{2018} \pmod{10} \\ &\equiv& (-1)^{2018}3^{2018} \pmod{10} \\ &\equiv& \mathbf{3^{2018} \pmod{10}} \\ &&&\boxed{{\color{blue}3^2} \pmod{10} \\ \equiv 9 \pmod{10} \\ \equiv 9-10 \pmod{10}\\ \equiv {\color{brown}-1} \pmod{10}} \\ &\equiv& \left({\color{blue}3^2}\right)^{1009} \pmod{10} \\ &\equiv& \left({\color{brown}-1}\right)^{1009} \pmod{10} \\ &\equiv& -1 \pmod{10} \\ &\equiv& -1+10 \pmod{10} \\ &\equiv& \mathbf{9 \pmod{10}} \\ \hline \end{array} \)

The unit digit of \(\mathbf{7^{2018}}\) is 9

What is unit digit of 7^2018

\(\begin{array}{|rcll|} \hline && \mathbf{{\color{red}7}^{2018} \pmod{10}}\\ &&&\boxed{{\color{red}7^2} \pmod{10} \\ \equiv 49 \pmod{10}\\ \equiv 9 \pmod{10} \\ \equiv 9-10 \pmod{10}\\ \equiv {\color{green}-1} \pmod{10}} \\ &\equiv& \left({\color{green}-1}\right)^{1009} \pmod{10} \\ &\equiv& -1 \pmod{10} \\ &\equiv& -1+10 \pmod{10} \\ &\equiv& \mathbf{9 \pmod{10}} \\ \hline \end{array} \)

The unit digit of \(\mathbf{7^{2018}}\) is 9