Question

The last digit of the number $(32{)}^{32}$ is

• A. 4
• B. 6
• C. 8
• D. none of these

Answer 1

Answer: (B)

$(32{)}^{32}=(2+3\times 10{)}^{32}$
$=2^{32}+10k,$ where $k\in N$
Therefore, last digits in $(32{)}^{32}=$ last digit in $(2{)}^{32}$
But $2^1=2,2^2=4,2^3=8,2^4=16,2^5=32$
$\therefore 2^{32}-{\left(2^5\right)}^6\cdot 2^2-(32{)}^6\cdot 4-(2+30{)}^6\cdot 4$
$=\left(2^6+10r\right)4,r\in N$
Last digit in $2^{32}=$ last digit in $(2{)}^6\cdot 4=$ last digit in $4\times 4=6$
$\therefore$ Last digit in $(32{)}^{32}=6$.