Question

Show that any number of the form 4^n,n
is natural
number can never
and with
the digit 0.​

Answer 1

Answer:

In 4n ends with 0 then it must have 5 as a factor.

But,4(n)=(22)n=22n show that 2 is the only prime factors of 4n.

Also we know from the fundamental theorem of arithnmetic that the prime factorisation of each numver is unique.

So, 5 is not a factor of 4n .

Hence, 4n can never end with the digit 0.