Question

What will be the unit digit of $(87{)}^{{75}^{{63}^{55}}}$?

Answer 1

Let $(87{)}^{{75}^{{63}^{55}}}=(87{)}^x$ where $x=(75{)}^{{63}^{55}}=(75{)}^{odd}$
$\because$ Cyclicity of $7$ is $4$,
$\therefore$ To find the last digit we have to find the remainder when $x$ is divided by $4$.
$x=(75{)}^{oddpower}=(76-1{)}^{oddpower}$
where, $x$ is divided by $4$ so, remainder will be $-1$ or $3$ but remainder should be positive always.
Therefore, the last digit is of $(87{)}^{{75}^{{63}^{55}}}$ will be $7^3=343$.
Hence, the last digit is of $(87{)}^{{75}^{{63}^{55}}}$ is $3$.