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

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!

Explore why the product 12n, for any natural number n, cannot end with the digits 0 or 5. This proof involves prime factorization, divisibility rules, and properties of numbers ending in specific digits. Suitable for students in Grades 9-12 studying number theory and divisibility.

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