Any number of the form ` 4^(n), n != N` can never end with the digit

Options

A. `0`
B. `4`
C. `6`
D. None

Answer

A

Explanation

In ` 4^(n)` ends with 0 then it must have 5 as a factor.
But,` 4(n) = (2^(2))^(n) = 2^(2n) ` show that 2 is the only prime factors of ` 4^(n)`.
Also we know from the fundamental theorem of arithnmetic that the prime factorisation of each numver is unieque.
So, 5 is not a factor of ` 4^(n)` .
Hence, ` 4 ^(n)` can never end with the digiy 0.