Show that one and only one out of n, n + 3, n + 6 or n + 9 is divisible by 4.

Answer

To show that one and only one out of the numbers \( n, n + 3, n + 6, \) or \( n + 9 \) is divisible by 4, we can analyze the possible values of \( n \) modulo 4.

### Step-by-Step Solution:

1. **Consider the possible remainders when \( n \) is divided by 4.**
- Any integer \( n \) can be expressed in one of the following forms based on its remainder when divided by 4:
- \( n \equiv 0 \mod 4 \)
- \( n \equiv 1 \mod 4 \)
- \( n \equiv 2 \mod 4 \)
- \( n \equiv 3 \mod 4 \)

2. **Analyze each case:**
- **Case 1:** If \( n \equiv 0 \mod 4 \)
- Then \( n \) is divisible by 4.
- \( n + 3 \equiv 3 \mod 4 \) (not divisible)
- \( n + 6 \equiv 2 \mod 4 \) (not divisible)
- \( n + 9 \equiv 1 \mod 4 \) (not divisible)
- **Conclusion:** Only \( n \) is divisible by 4.

- **Case 2:** If \( n \equiv 1 \mod 4 \)
- Then \( n + 3 \equiv 0 \mod 4 \) (divisible)
- \( n \equiv 1 \mod 4 \) (not divisible)
- \( n + 6 \equiv 2 \mod 4 \) (not divisible)
- \( n + 9 \equiv 3 \mod 4 \) (not divisible)
- **Conclusion:** Only \( n + 3 \) is divisible by 4.

- **Case 3:** If \( n \equiv 2 \mod 4 \)
- Then \( n + 6 \equiv 0 \mod 4 \) (divisible)
- \( n \equiv 2 \mod 4 \) (not divisible)
- \( n + 3 \equiv 1 \mod 4 \) (not divisible)
- \( n + 9 \equiv 3 \mod 4 \) (not divisible)
- **Conclusion:** Only \( n + 6 \) is divisible by 4.

- **Case 4:** If \( n \equiv 3 \mod 4 \)
- Then \( n + 9 \equiv 0 \mod 4 \) (divisible)
- \( n \equiv 3 \mod 4 \) (not divisible)
- \( n + 3 \equiv 0 \mod 4 \) (not divisible)
- \( n + 6 \equiv 2 \mod 4 \) (not divisible)
- **Conclusion:** Only \( n + 9 \) is divisible by 4.

3. **Final Conclusion:**
- In each case, we find that exactly one of the numbers \( n, n + 3, n + 6, n + 9 \) is divisible by 4. Therefore, we have shown that one and only one of these numbers is divisible by 4.