Question

Show that 12ⁿ cannot end with the digit 0 or 5 for any natural number n.

Answer 1

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 ${12}^n$ contains the prime number 5.

Now, 12 = 2 × 2 × 3 = 2² × 3
⇒ 12ⁿ = (2² × 3)ⁿ = 2²ⁿ × $3^n$ [since, there is no term contains 5]
Hence, there is no value of n for which 12ⁿ ends with digit zero or five.