Question

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

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

Answer 1

Answer: (B)

$6$

${\left(32\right)}^{32}={(2+3\times 10)}^{32}={\left(2\right)}^{32}+10k,$ where $k\in N$

Therefore, last digits in ${\left(32\right)}^{32}=$ last digit in ${\left(2\right)}^{32}$

But $2^1=2,2^2=4,2^3=8,2^4=16,2^5=32$

$\therefore {\left(2\right)}^{32}={\left(2^5\right)}^6\times 2^2={\left(32\right)}^6\times 4={(30+2)}^6\times 4$

$=(2^6+10r)\times 4,r\in N$

Last digit in ${\left(2\right)}^{32}=$ last digit in ${\left(2\right)}^6\times 4=$ last digit in $4\times 4=6$

$\therefore$ Last digit is ${\left(32\right)}^{32}=6$