Check whether $6^n$ can end with the digit 0 for any natural number n. [2 MARKS]

Concept : 1 Mark
Application : 1 Mark

Any number ending with 0 is divisible by 5 and has 5 as one of its prime factors

But $6^n=(2\times 3{)}^n$

$\Rightarrow$The $6^n$ does not have 5 as one of its prime factor.

By uniqueness of the fundamental theorem of arithmetic, there is no other primes in the prime factorisation of $6^n$

$\therefore$ $6^n$ cannot end with 0 for any natural number n.