Question

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

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

Answer 1

The correct option is B 6

$(32{)}^{32}=(2^5{)}^{32}=2^{160}$

Therefore $(32{)}^{3}2$ can be rewritten as $2^{160}$

The exponent of 2 has a regular pattern:

$2^1=2$ …………………(i)

$2^2=4$ …………………(ii)

$2^3=8$ …………………(iii)

$2^4=16$ (last digit = 6) …………………(iV)

And then it repeats

Find the last digit of ${32}^{32}$

$160÷4=40$

Then, It will fall into eq (iv)

Hence, the last digit is $6$