Show that `12^(th)` cannot end with the digit 0 or 5 for any natural number n.

Explanation

Here `" " 12 = 2 xx 2 xx 3`
`= 2^(2) xx 3`
`rArr" " 12^(n) = (2^(2) xx 3)^(n) = 2^(2n) xx 3^(n)`
We know that if any numbers ends with the digits 0 or 5, its is always divisible by 5.
But the prime factorisation of `12^(n)` does not contain the prime number 5.
Hence, `12^(n)` cannot end with the digits 0 or 5 for any natural number n.