Question

Find the last digit of the number ${\left( {32} \right)^{32}}$.
A. 4
B. 6
C. 8
D. None of these

Answer 1

Hint: Units digit of ${(xyz)^{abc}}$ is the same as units digit of ${(z)^{abc}}$.

Complete step-by-step answer:
We have to find the units digit of ${\left( {32} \right)^{32}}$. It can be written as
${\left( {32} \right)^{32}} = {\left( {2 + (3 \times 10)} \right)^{32}} = {2^{32}} + 10k,$ where $k \in N$
Therefore, last digits in ${\left( {32} \right)^{32}}$= last digit in ${\left( 2 \right)^{32}}$
Let us observe the cyclicity of 2.
${2^1} = 2,{2^2} = 4,{2^3} = 8,{2^4} = 16,{2^5} = 32,{2^6} = 64...$
Observe the sequence of unit digits for different powers of 2.
$\therefore {\left( 2 \right)^{32}} = {\left( {{2^5}} \right)^6} \times {2^2} = {(32)^6} \times 4 = {(30 + 2)^6} \times 4$
$ = ({2^6} + 10r) \times 4,$ where $r \in N$
Last digit in ${\left( 2 \right)^{32}}$ = last digit in ${(2)^6} \times 4$
The value of ${\left( 2 \right)^6}$=64 and
The units digit of ${\left( 2 \right)^6}$is 4
= Last digit in $4 \times 4 = 6$
$\therefore $ Last digit of ${(32)^{32}}$ is 6.

Note: Unit’s digit of a number is the digit in one’s place of the number. i.e., the rightmost digit of the number. We are finding the unit’s digit of numbers with large powers. We can also use remainder of power when divided with cyclicity of the number in the unit's place. In the above problem we have 2 in the unit’s place.
The cyclicity of 2 is 4. That means the unit’s digits of ${2^x}\left[ {x > 0} \right]$ follows the pattern of digits 2, 4, 8 and 6. The power in the problem is 32. When we divide the power by cyclicity, we get a remainder
$ \Rightarrow \dfrac{{32}}{4}$ Remainder = 0.
Remainder 0 represents that we have to take the last number in cyclicity pattern which becomes the required answer = 6.