Question

Prove that there is no natural number for which $4^n$ ends with the digit zero.

Answer 1

We know that any positive integer ending with the digit zero is divisible by $5$ and so its prime factorization must contain the prime $5$
We have
$4^n=2^{2n}$
$\Rightarrow$ The only prime in the factorization of $4^n$ is $2$.
$\Rightarrow$ There is no other primes in the factorization of $4^n=2^{2n}$
[By uniqueness of the Fundamental theorem of Arithmetic]
$\Rightarrow$ $5$ does not occur in the prime factorization of $4^n$ for any $n$.
$\Rightarrow$ $4^n$ does not end with the digit zero for any natural $n$.