Question

$6^n$ can end with the digit $0$ for any natural number.
If true then enter $1$ and if false then enter $0$

Answer 1

Any positive integer ending with the digit zero is divisible by the prime number $5$ and so its prime factorizations must contain the prime $5$

Now, lets take $6^n$
$6^n=(2\times 3{)}^n=2^n\times 3^n$
$\Rightarrow$ The prime in the factorization of $6^n$ is $2$ and $3$
$\Rightarrow$ $5$ does not occur in the prime factorization of $6^n$ for any $n$.
$\Rightarrow 6^n$ does not end with the digit zero for any natural number $n$.

So, the given statement is False.

So, enter $0$.