The number of ordered pairs (m,n)`(m,n epsilon {1,2,……..20})` such that `3^(m)+7^(n)` is a multiple of 10, is

Options

A. 100
B. 200
C. `4!xx4!`
D. None of these

Answer

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 can follow these steps:

### Step 1: Determine the last digits of \(3^m\) and \(7^n\)

The last digits of powers of 3 and 7 repeat in cycles:
- For \(3^m\):
- \(3^1 \equiv 3\) (last digit 3)
- \(3^2 \equiv 9\) (last digit 9)
- \(3^3 \equiv 7\) (last digit 7)
- \(3^4 \equiv 1\) (last digit 1)
- This cycle repeats every 4 terms: \(3, 9, 7, 1\).

- For \(7^n\):
- \(7^1 \equiv 7\) (last digit 7)
- \(7^2 \equiv 9\) (last digit 9)
- \(7^3 \equiv 3\) (last digit 3)
- \(7^4 \equiv 1\) (last digit 1)
- This cycle also repeats every 4 terms: \(7, 9, 3, 1\).

### Step 2: Identify the conditions for \(3^m + 7^n\) to be a multiple of 10

We need to find combinations of the last digits of \(3^m\) and \(7^n\) such that their sum is congruent to 0 modulo 10. The possible last digits of \(3^m\) and \(7^n\) are:
- Last digits of \(3^m\): \(3, 9, 7, 1\)
- Last digits of \(7^n\): \(7, 9, 3, 1\)

Now we check the combinations:
- If \(3^m \equiv 3\), then \(7^n\) must be \(7\) (since \(3 + 7 = 10\)).
- If \(3^m \equiv 9\), then \(7^n\) must be \(1\) (since \(9 + 1 = 10\)).
- If \(3^m \equiv 7\), then \(7^n\) must be \(3\) (since \(7 + 3 = 10\)).
- If \(3^m \equiv 1\), then \(7^n\) must be \(9\) (since \(1 + 9 = 10\)).

### Step 3: Count the occurrences of each last digit

Since \(m\) and \(n\) can take values from 1 to 20, we can determine how many times each last digit occurs:
- For \(3^m\):
- Last digit 3 occurs for \(m \equiv 1 \mod 4\) → \(m = 1, 5, 9, 13, 17\) (5 values)
- Last digit 9 occurs for \(m \equiv 2 \mod 4\) → \(m = 2, 6, 10, 14, 18\) (5 values)
- Last digit 7 occurs for \(m \equiv 3 \mod 4\) → \(m = 3, 7, 11, 15, 19\) (5 values)
- Last digit 1 occurs for \(m \equiv 0 \mod 4\) → \(m = 4, 8, 12, 16, 20\) (5 values)

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

### Step 4: Calculate the total number of valid pairs

Now we can calculate the total number of valid pairs \((m, n)\):
1. For \(3^m \equiv 3\) and \(7^n \equiv 7\): \(5 \times 5 = 25\)
2. For \(3^m \equiv 9\) and \(7^n \equiv 1\): \(5 \times 5 = 25\)
3. For \(3^m \equiv 7\) and \(7^n \equiv 3\): \(5 \times 5 = 25\)
4. For \(3^m \equiv 1\) and \(7^n \equiv 9\): \(5 \times 5 = 25\)

Adding these together gives:
\[ 25 + 25 + 25 + 25 = 100 \]

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

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