Question

The unit digit in the expression
$\left({36}^{234}\right)\left({33}^{512}\right)\left({39}^{180}\right)-\left({54}^{29}\right)\left({25}^{123}\right)\left({31}^{512}\right)$ will be

• A. $8$
• B. $0$
• C. $6$
• D. $5$

Answer 1

The correct option is C $6$

${36}^{234}$ since 6 is unit digit and any power to 6 will be 6

$\Rightarrow$ ${33}^{512}$ here 3 is unit digit so take last two terms of power i.e 12 and divide by 4 (because 3 has cyclicity of 3,9,7,1). So $12/4$ i.e reminder 0 so unit digit will be $1.$

$\Rightarrow$ ${39}^{180}$ here 9 is unit digit same process repeat as above did. $80/2$ (here divide by 2 because 9 has only 2 cycle 9 & 1 rem 0 ) so unit digit $1.$

$\Rightarrow$ ${54}^{29}$ - unit digit $4.$

$\Rightarrow$ ${25}^{123}$ - any power to 5 will be $5.$

$\Rightarrow$ ${31}^{512}$ - any power to 1 will be $1.$

$\Rightarrow$ So, $(6\times 1\times 1)-(4\times 5\times 1)=06-20$

$\Rightarrow$ Subtract unit digit = $6-0=6$

$\therefore$ The unit digit in given expression will be $6$.