To prove that $12^n$ cannot end with the digit 0 or 5 for any natural number $n$, we need to examine the last digit of powers of 12.

Step 1: Analyze Powers of 12 Modulo 10

First, let's consider the last digit of $12^n$. The last digit of a number is the same as the remainder when the number is divided by 10. This is equivalent to calculating $12^n \mod 10$.

Step 2: Compute the Last Digit of Powers of 12

Let's calculate the first few powers of 12 and observe the pattern in the last digit:

• $12^1 = 12$
  Last digit = $12 \mod 10 = 2$
• $12^2 = 144$
  Last digit = $144 \mod 10 = 4$
• $12^3 = 1728$
  Last digit = $1728 \mod 10 = 8$
• $12^4 = 20736$
  Last digit = $20736 \mod 10 = 6$
• $12^5 = 248832$
  Last digit = $248832 \mod 10 = 2$

The sequence of last digits for $12^n$ is 2, 4, 8, 6, and then it repeats every four terms: 2, 4, 8, 6.

Step 3: Identify the Pattern

The pattern for the last digit of $12^n$ is as follows:

• $n \equiv 1 \pmod{4}$ leads to a last digit of 2.
• $n \equiv 2 \pmod{4}$ leads to a last digit of 4.
• $n \equiv 3 \pmod{4}$ leads to a last digit of 8.
• $n \equiv 0 \pmod{4}$ leads to a last digit of 6.

Step 4: Conclusion

• The last digit of $12^n$ can only be 2, 4, 8, or 6.
• Therefore, $12^n$ cannot end with the digit 0 or 5 for any natural number $n$.

This completes the proof.

Would you like any further details on this proof?

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Here are 5 related questions:

1. How would you prove that $2^n$ cannot end with the digit 5 for any natural number $n$?
2. What is the last digit of $7^n$ for different values of $n$?
3. Can you generalize a method to find the last digit of any power $a^n$?
4. Why is it impossible for $3^n$ to end in the digit 0 for any $n$?
5. How does modular arithmetic help in understanding the properties of last digits in powers?

Tip: When dealing with last digits, think in terms of modular arithmetic (mod 10). It simplifies the problem to just examining the remainder when divided by 10.