Find the number of ordered pairs `(m,n)epsilon {1,2,…..20}` such that `3^(m)+7^(n)` is divisible by 10.

Answer

To solve the problem of finding the number of ordered pairs \((m, n)\) such that \(3^m + 7^n\) is divisible by 10, we need to analyze the last digits of \(3^m\) and \(7^n\).

### Step-by-Step Solution:

1. **Identify the last digits of \(3^m\)**:
- The last digits of powers of 3 cycle every 4:
- \(3^1 \equiv 3\)
- \(3^2 \equiv 9\)
- \(3^3 \equiv 7\)
- \(3^4 \equiv 1\)
- Thus, the last digits of \(3^m\) are:
- \(m \equiv 1 \mod 4 \rightarrow 3\)
- \(m \equiv 2 \mod 4 \rightarrow 9\)
- \(m \equiv 3 \mod 4 \rightarrow 7\)
- \(m \equiv 0 \mod 4 \rightarrow 1\)

2. **Identify the last digits of \(7^n\)**:
- The last digits of powers of 7 also cycle every 4:
- \(7^1 \equiv 7\)
- \(7^2 \equiv 9\)
- \(7^3 \equiv 3\)
- \(7^4 \equiv 1\)
- Thus, the last digits of \(7^n\) are:
- \(n \equiv 1 \mod 4 \rightarrow 7\)
- \(n \equiv 2 \mod 4 \rightarrow 9\)
- \(n \equiv 3 \mod 4 \rightarrow 3\)
- \(n \equiv 0 \mod 4 \rightarrow 1\)

3. **Determine combinations for divisibility by 10**:
- For \(3^m + 7^n\) to be divisible by 10, the last digits must add up to 10:
- \(3 + 7 = 10\)
- \(9 + 1 = 10\)
- \(7 + 3 = 10\)
- \(1 + 9 = 10\)

4. **Count valid pairs**:
- **Case 1**: \(3^m \equiv 3\) and \(7^n \equiv 7\)
- \(m \equiv 1 \mod 4\): Possible values are \(1, 5, 9, 13, 17\) (5 values)
- \(n \equiv 1 \mod 4\): Possible values are \(1, 5, 9, 13, 17\) (5 values)
- Total pairs = \(5 \times 5 = 25\)

- **Case 2**: \(3^m \equiv 9\) and \(7^n \equiv 1\)
- \(m \equiv 2 \mod 4\): Possible values are \(2, 6, 10, 14, 18\) (5 values)
- \(n \equiv 0 \mod 4\): Possible values are \(4, 8, 12, 16, 20\) (5 values)
- Total pairs = \(5 \times 5 = 25\)

- **Case 3**: \(3^m \equiv 7\) and \(7^n \equiv 3\)
- \(m \equiv 3 \mod 4\): Possible values are \(3, 7, 11, 15, 19\) (5 values)
- \(n \equiv 3 \mod 4\): Possible values are \(3, 7, 11, 15, 19\) (5 values)
- Total pairs = \(5 \times 5 = 25\)

- **Case 4**: \(3^m \equiv 1\) and \(7^n \equiv 9\)
- \(m \equiv 0 \mod 4\): Possible values are \(4, 8, 12, 16, 20\) (5 values)
- \(n \equiv 2 \mod 4\): Possible values are \(2, 6, 10, 14, 18\) (5 values)
- Total pairs = \(5 \times 5 = 25\)

5. **Calculate the total number of pairs**:
- Total pairs from all cases = \(25 + 25 + 25 + 25 = 100\)

### Final Answer:
The total number of ordered pairs \((m, n)\) such that \(3^m + 7^n\) is divisible by 10 is **100**.