To solve the problem of finding the number of ordered pairs \((m, n)\) such that \(3^m + 7^n\) is a multiple of 10, we need to analyze the last digits of \(3^m\) and \(7^n\).
### Step-by-Step Solution:
1. **Identify Last Digits of Powers of 3 and 7:**
- The last digits of \(3^m\) cycle through:
- \(3^1 \equiv 3\)
- \(3^2 \equiv 9\)
- \(3^3 \equiv 7\)
- \(3^4 \equiv 1\)
- This cycle repeats every 4 terms: \(3, 9, 7, 1\).
- The last digits of \(7^n\) cycle through:
- \(7^1 \equiv 7\)
- \(7^2 \equiv 9\)
- \(7^3 \equiv 3\)
- \(7^4 \equiv 1\)
- This cycle also repeats every 4 terms: \(7, 9, 3, 1\).
2. **Determine Conditions for Divisibility by 10:**
- For \(3^m + 7^n\) to be divisible by 10, the last digits must sum to 10 or 0.
- The possible pairs of last digits that satisfy this condition are:
- \(3 + 7 = 10\)
- \(9 + 1 = 10\)
- \(7 + 3 = 10\)
- \(1 + 9 = 10\)
3. **Find Values of \(m\) for Each Last Digit:**
- **Last Digit 9 for \(3^m\):**
- \(m \equiv 2 \mod 4\) (i.e., \(m = 2, 6, 10, 14, 18\)) → 5 values.
- **Last Digit 1 for \(7^n\):**
- \(n \equiv 0 \mod 4\) (i.e., \(n = 4, 8, 12, 16, 20\)) → 5 values.
- **Last Digit 1 for \(3^m\):**
- \(m \equiv 0 \mod 4\) (i.e., \(m = 4, 8, 12, 16, 20\)) → 5 values.
- **Last Digit 9 for \(7^n\):**
- \(n \equiv 2 \mod 4\) (i.e., \(n = 2, 6, 10, 14, 18\)) → 5 values.
4. **Count the Ordered Pairs:**
- For the pairs \( (3, 7) \) and \( (9, 1) \):
- \(5 \times 5 = 25\) pairs.
- For the pairs \( (1, 9) \) and \( (7, 3) \):
- \(5 \times 5 = 25\) pairs.
- Total pairs from both cases:
- \(25 + 25 = 50\) pairs.
5. **Final Calculation:**
- Since there are two distinct cases (where \(3^m\) ends with 9 and \(7^n\) ends with 1, and vice versa), we multiply the number of pairs by 2:
- Total pairs = \(25 + 25 = 50\) from both cases.
- Thus, the total number of ordered pairs \((m, n)\) such that \(3^m + 7^n\) is divisible by 10 is \(50\).
### Final Answer:
The number of ordered pairs \((m, n)\) such that \(3^m + 7^n\) is a multiple of 10 is **50**.
