Show that 12n cannot end with the digit 0 or 5 for any natural number n.
Cannot end with 0 or 5.
Step 1: Prime factorisation of \(12^n\).
We know that
\(12^n = (3 \times 4)^n\).
This can be written as
\(12^n = 3^n \times 4^n\).
And since \(4^n = (2^2)^n = 2^{2n}\), we finally get
\(12^n = 3^n \times 2^{2n}\).
So \(12^n\) has only the prime factors 2 and 3.
Step 2: Condition for a number to end with 0 or 5.
If a number ends with digit 0, it must be divisible by 10. That means it must have both 2 and 5 as prime factors.
If a number ends with digit 5, it must be divisible by 5. So it must contain 5 as a prime factor.
Step 3: Check divisibility by 5.
From Step 1, we saw that \(12^n\) contains only factors of 2 and 3.
It does not have any factor of 5.
Conclusion.
Since \(12^n\) has no factor of 5, it can never end with 0 or 5.
Therefore, for any natural number \(n\), the number \(12^n\) cannot end with the digit 0 or 5.