Question 5
Check whether $6^n$ can end with the digit 0 for any natural number n.

If any number ends with a digit 0, it should be divisible by 10.
In other words, it will also be divisible by 2 and 5 as $10=2\times 5$
Now, $6^n=(2\times 3{)}^n$
It can be seen that 5 does not appear in the prime factorisation of $6^n$.
Hence, for any value of n, $6^n$ will not be divisible by 5, and thereby, $6^n$ will not be divisible by 10,
Therefore, $6^n$ cannot end with the digit 0 for any natural number n.