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The cyclicity of the digits are as given below


0,1,5 and 6 have a cyclicity of 1

4 and 9 (the perfect squares) have a cyclicity of 2

All others i.e 2, 3, 7 and 8 have a cyclicity of 4.


To Find the units place of \(a^b \), where a is the base and b is the power.


Divide the power by the cyclicity of the base and find the remainder.


Here 2 cases arise.


Case 1: If the remainder = 0


For base ending in 2, 4 or 8 the last digit will be 6

For base ending in 3, 7 or 9 the last digit will be 1


Case 2: If the remainder \(\neq 0\), then the last digit will be the base to the power of the remainder.




In the question given above: To find the units place of \(2^{39}\)

The base is 2 whose cyclicity is 4 and the power is 39


R\([\frac{39}{4}]\) = 3

The base raised to the power gives the answer = \(2^3\) = 8


Option D


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We need to find the units digit of \(2^{39}\)

Lets start by finding the cyclicity of units' digit in powers of 2

\(2^1\) units’ digit is 2
\(2^2\) units’ digit is 4
\(2^3\) units’ digit is 8
\(2^4\) units’ digit is 6
\(2^5\) units’ digit is 2

That means that units digit of power of 2 has a cycle of 4

=> We need to divide the power (39) by 4 and check what is the remainder
39 divided by 4 gives 3 remainder

=> Units digit of \(2^{39}\) = Units digit of \(2^3\) = 8

So, Answer will be D
Hope it helps!

Link to Theory for Last Two digits of exponents here.

Link to Theory for Units' digit of exponents here.
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