Answer :

If any number ends with the digit 0 or 5, it is always divisible by 5. If 12n ends with the digit zero it must be divisible by 5. This is possible only if prime factorisation of 12n contains the prime number 5.

12 = 2 × 2 × 3 = 2^{2} × 3

⇒ 12n = (2^{2} × 3)n = 2^{2}n × 3n

Since its prime factorisation does not contain 5.

Hence, 12n cannot end with the digit 0 or 5 for any natural number n.

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