Question

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

Answer 1

Let $(87{)}^{{75}^{{63}^{55}}}$ = $(87{)}^x$
where x = $(75{)}^{{63}^{55}}$ = $(87{)}^{odd}$
$\because$ 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 always positive
Therefore the last digit of $(87{)}^{{75}^{{63}^{55}}}$ will be $(7{)}^3=343$
Hence the last digit is of $(87{)}^{{75}^{{63}^{55}}}$ is 3