Question

One’s digit of the cube of 54 is:
A) 6
B) 4
C) 2
D) 9

Answer 1

Hint: Cube of a number means that the three times multiplication of the number to itself.
For example, Let the number = a
Cube of a $\mathop { = {\rm{ a}}}\nolimits^3 {\rm{ = a}} \times {\rm{a}} \times {\rm{a}}$
One’s place or unit digit is the leftmost digit of the number.
The power cycle of a number: the numbers have cyclicity (repetition) of their units’ digits for increasing powers.
For example:
 $\begin{array}{l}
\mathop 2\nolimits^1 {\rm{ }} = {\rm{ }}2,{\rm{ unit digit = 2}}\\
\mathop 2\nolimits^2 {\rm{ }} = {\rm{ }}4,{\rm{ unit digit = 4}}\\
\mathop 2\nolimits^3 {\rm{ }} = {\rm{ }}8,{\rm{ unit digit = 8}}\\
\mathop 2\nolimits^4 {\rm{ }} = 16,{\rm{ unit digit = 6}}\\
\mathop 2\nolimits^5 {\rm{ }} = 32,{\rm{ unit digit = 2}}
\end{array}$
The power cycle of 2 = (2, 4, 8, 6)
The power cycle of 2 contains four numbers. Therefore, its cyclicity is 4.
Example question on use of cyclicity: find the unit place of $\mathop 2\nolimits^{251} $ .
sol: The power cycle of 2 = (2, 4, 8, 6)
                 cyclicity of 2 = 4,
                 On dividing the power of 2 by cyclicity
                251 divided by 4, remainder is 1.
                 $\mathop 2\nolimits^{251} {\rm{ reduced to }}\mathop 2\nolimits^1 $
                $\mathop 2\nolimits^1 = 2$, unit place = 2

Complete step by step solution:
Step 1
Cube of 54 $\begin{array}{l}
\mathop { = {\rm{ }}54}\nolimits^3 \\
 = 54 \times 54 \times 54
\end{array}$

Step 2
Unit place of 54 = 4 …… (1)

Step 3
The power cycle of 4:
$\begin{array}{l}
\mathop 4\nolimits^1 {\rm{ }} = {\rm{ 4}},{\rm{ unit digit = 4}}\\
\mathop 4\nolimits^2 {\rm{ }} = {\rm{ 16}},{\rm{ unit digit = 6}}\\
\mathop 4\nolimits^3 {\rm{ }} = {\rm{ 64}},{\rm{ unit digit = 4}}\\
\mathop 4\nolimits^4 {\rm{ }} = 256,{\rm{ unit digit = 6}}
\end{array}$
Therefore, the power cycle of 4 = (4, 6)

Step 4
The power cycle of 4 contains two numbers. Therefore, its cyclicity is 2. …… (2)

Step 5
Unit place of $\mathop {54}\nolimits^3 $= unit place of $\mathop 4\nolimits^3 $

Step 6
Dividing the power of 54 by cyclicity of 4
3 is divided by 2, remainder = 1 ( cyclicity of 4 is 2)
$\mathop 4\nolimits^3 {\rm{ reduced to }}\mathop 4\nolimits^1 $
Unit place of $\mathop 4\nolimits^1 $= 4

Step 7
Unit place of $\mathop {54}\nolimits^3 $= unit place of $\mathop 4\nolimits^3 $ = Unit place of $\mathop 4\nolimits^1 $= 4 …… (3)
$ \Rightarrow $ Unit place of $\mathop {54}\nolimits^3 $ = 4 (from (3))

Unit place of $\mathop {54}\nolimits^3 = 4$ . The correct option is (B)

Note:
In the Indian numeral system, the place values of digits go in the sequence of Ones, Tens, Hundreds, Thousands, Ten Thousand, Lakhs, Ten Lakhs, Crores and so on.
For example: write the place value of all the digits of the number: 2,45,34,720.

Place value	Digit
Ones	0
Tens	2
Hundreds	7
Thousands	4
Ten-thousands	3
Lakhs	5
Ten-lakhs	4
Crore	2

Power cycle of 3:
$\begin{array}{l}
\mathop 3\nolimits^1 {\rm{ }} = {\rm{ 3}},{\rm{ unit digit = 3}}\\
\mathop 3\nolimits^2 {\rm{ }} = {\rm{ 9}},{\rm{ unit digit = 9}}\\
\mathop 3\nolimits^3 {\rm{ }} = {\rm{ 27}},{\rm{ unit digit = 7}}\\
\mathop 3\nolimits^4 {\rm{ }} = {\rm{ 81}},{\rm{ unit digit = 1}}\\
\mathop 3\nolimits^5 {\rm{ }} = 243,{\rm{ unit digit = 3}}
\end{array}$
The power cycle of 3 = (3, 9, 7, 1)
The power cycle of 3 contains four numbers. Therefore, its cyclicity is 4.
Power cycle of 5:
$\begin{array}{l}
\mathop 5\nolimits^1 {\rm{ }} = {\rm{ 5}},{\rm{ unit digit = 5}}\\
\mathop 5\nolimits^2 {\rm{ }} = {\rm{ 25}},{\rm{ unit digit = 5}}\\
\mathop 5\nolimits^3 {\rm{ }} = {\rm{ 125}},{\rm{ unit digit = 5}}\\
\mathop 5\nolimits^4 {\rm{ }} = 625,{\rm{ unit digit = 5}}
\end{array}$
The power cycle of 5 = ( 5 )
The power cycle of 5 contains only one number, i.e. 5. Therefore, no matter what is the power of 5, the unit digit will always be 5.