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

Answer

To find 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 1: Determine the last digits of \(3^m\)
The last digits of \(3^m\) follow a pattern:
- \(3^1 \equiv 3\)
- \(3^2 \equiv 9\)
- \(3^3 \equiv 7\)
- \(3^4 \equiv 1\)

This pattern repeats every 4 terms:
- Last digits: \(3, 9, 7, 1\)

Thus, for \(m\) values:
- \(m \equiv 1 \mod 4\) gives last digit \(3\)
- \(m \equiv 2 \mod 4\) gives last digit \(9\)
- \(m \equiv 3 \mod 4\) gives last digit \(7\)
- \(m \equiv 0 \mod 4\) gives last digit \(1\)

### Step 2: Determine the last digits of \(7^n\)
The last digits of \(7^n\) also follow a pattern:
- \(7^1 \equiv 7\)
- \(7^2 \equiv 9\)
- \(7^3 \equiv 3\)
- \(7^4 \equiv 1\)

This pattern also repeats every 4 terms:
- Last digits: \(7, 9, 3, 1\)

Thus, for \(n\) values:
- \(n \equiv 1 \mod 4\) gives last digit \(7\)
- \(n \equiv 2 \mod 4\) gives last digit \(9\)
- \(n \equiv 3 \mod 4\) gives last digit \(3\)
- \(n \equiv 0 \mod 4\) gives last digit \(1\)

### Step 3: Find combinations where \(3^m + 7^n\) is divisible by 10
For \(3^m + 7^n\) to be divisible by 10, the last digits must sum to 10. The valid combinations are:
1. Last digit of \(3^m\) is \(9\) and last digit of \(7^n\) is \(1\).
2. Last digit of \(3^m\) is \(1\) and last digit of \(7^n\) is \(9\).
3. Last digit of \(3^m\) is \(7\) and last digit of \(7^n\) is \(3\).
4. Last digit of \(3^m\) is \(3\) and last digit of \(7^n\) is \(7\).

### Step 4: Count the valid \(m\) and \(n\) values
- For \(3^m\):
- \(m \equiv 2 \mod 4\) (last digit \(9\)): \(m = 2, 6, 10, 14, 18\) → 5 options
- \(m \equiv 0 \mod 4\) (last digit \(1\)): \(m = 4, 8, 12, 16, 20\) → 5 options
- \(m \equiv 3 \mod 4\) (last digit \(7\)): \(m = 3, 7, 11, 15, 19\) → 5 options
- \(m \equiv 1 \mod 4\) (last digit \(3\)): \(m = 1, 5, 9, 13, 17\) → 5 options

- For \(7^n\):
- \(n \equiv 0 \mod 4\) (last digit \(1\)): \(n = 4, 8, 12, 16, 20\) → 5 options
- \(n \equiv 2 \mod 4\) (last digit \(9\)): \(n = 2, 6, 10, 14, 18\) → 5 options
- \(n \equiv 3 \mod 4\) (last digit \(3\)): \(n = 3, 7, 11, 15, 19\) → 5 options
- \(n \equiv 1 \mod 4\) (last digit \(7\)): \(n = 1, 5, 9, 13, 17\) → 5 options

### Step 5: Calculate the total combinations
1. For \(3^m\) ends with \(9\) and \(7^n\) ends with \(1\): \(5 \times 5 = 25\)
2. For \(3^m\) ends with \(1\) and \(7^n\) ends with \(9\): \(5 \times 5 = 25\)
3. For \(3^m\) ends with \(7\) and \(7^n\) ends with \(3\): \(5 \times 5 = 25\)
4. For \(3^m\) ends with \(3\) and \(7^n\) ends with \(7\): \(5 \times 5 = 25\)

### Final Count
Total ordered pairs \((m,n)\) = \(25 + 25 + 25 + 25 = 100\).

Thus, the number of ordered pairs \((m,n)\) such that \(3^m + 7^n\) is divisible by 10 is **100**.