To show that $12n$ cannot end with the digit 0 or 5 for any natural number $n$, we will explore the properties of the number 12 and the digits 0 and 5.

Step-by-Step Reasoning:

1. Prime Factorization of 12: The number 12 can be factored as: $12 = 2^2 \times 3$ So, $12n$ is always a multiple of both 2 and 3, for any natural number $n$.

2. Properties of Numbers Ending in 0 or 5:

   • A number ending in 0 is divisible by 10. This implies that it must be divisible by both 2 and 5 (since $10 = 2 \times 5$).
   • A number ending in 5 must be divisible by 5 (because numbers ending in 5 are multiples of 5).

3. Requirement for Ending in 0 or 5:

   • For $12n$ to end in 0, it must be divisible by both 2 and 5.
   • For $12n$ to end in 5, it must be divisible by 5.

4. Divisibility by 5: The prime factorization of 12 is $2^2 \times 3$. Notice that there is no factor of 5 in the factorization of 12. Therefore, regardless of the value of $n$, $12n$ can never have a factor of 5 unless $n$ itself has a factor of 5.

5. Contradiction for Ending in 0 or 5:

   • Since $12n$ does not include a factor of 5, it cannot be divisible by 5.
   • This means that $12n$ cannot end in either 0 or 5, because both cases require divisibility by 5, which $12n$ lacks.

Conclusion:

For any natural number $n$, $12n$ cannot end with the digit 0 or 5 because it is not divisible by 5. Therefore, the digits 0 or 5 cannot appear as the last digit of $12n$.

Would you like further details or have any questions?

Here are 5 related questions to deepen your understanding:

1. Can a number with prime factors other than 5 ever end in 0 or 5?
2. How does the divisibility rule for 10 affect the possible last digits of a number?
3. What is the smallest natural number $n$ such that $12n$ ends with a different specific digit?
4. Can a number with prime factorization involving only 2s and 3s end in any digit other than 0 or 5?
5. How would this proof change if we asked about a different number instead of 12?

Tip: Always examine the prime factorization of a number to determine whether it can have a specific last digit!