Show that `12^n` cannot end with the digits `0` or `5` for any natural number `n`

Explanation

If any number ends with the digit 0 or 5, it is always divisible by 5.
This is possible only if prime factorisation of `12^(n)` contains the prime number 5.
Now, `12=2 xx 2 xx 3=2^(2)xx3`
`Rightarrow 12^(m)=(2^(2)xx3)^(n) =2^(2n)xx3^(m)` [Since, there is no term contains 5]
Hence, there is no value of n `in` N for which `12^(n)` ends with digit zero or live.