The number of ordered pairs `(m,n)` where `m`, `n in {1,2,3,…,50}`, such that `6^(m)+9^(n)` is a multiple of `5` is

Options

A. `1250`
B. `2500`
C. `625`
D. `500`

Explanation

To solve the problem of finding the number of ordered pairs \((m,n)\) such that \(6^m + 9^n\) is a multiple of \(5\), we can follow these steps:

### Step 1: Determine the units digit of \(6^m\)
The units digit of \(6^m\) for any positive integer \(m\) is always \(6\). This is because:
- \(6^1 = 6\)
- \(6^2 = 36\) (units digit is \(6\))
- \(6^3 = 216\) (units digit is \(6\))
- And so on...

Thus, for any \(m\), the units digit of \(6^m\) is \(6\).

### Step 2: Determine the units digit of \(9^n\)
The units digit of \(9^n\) alternates based on whether \(n\) is odd or even:
- If \(n\) is odd: \(9^1 = 9\), \(9^3 = 729\) (units digit is \(9\))
- If \(n\) is even: \(9^2 = 81\), \(9^4 = 6561\) (units digit is \(1\))

So, we have:
- For odd \(n\): units digit of \(9^n\) is \(9\)
- For even \(n\): units digit of \(9^n\) is \(1\)

### Step 3: Analyze the sum \(6^m + 9^n\)
We need \(6^m + 9^n\) to be a multiple of \(5\). This means we need to analyze the possible sums based on the units digits we found.

1. **When \(n\) is odd**:
- Units digit of \(6^m + 9^n = 6 + 9 = 15\) (which is a multiple of \(5\))

2. **When \(n\) is even**:
- Units digit of \(6^m + 9^n = 6 + 1 = 7\) (which is not a multiple of \(5\))

### Step 4: Count the valid pairs \((m,n)\)
From the analysis:
- \(m\) can take any value from \(1\) to \(50\), giving us \(50\) options.
- \(n\) must be odd to satisfy the condition. The odd numbers from \(1\) to \(50\) are \(1, 3, 5, \ldots, 49\). This is an arithmetic sequence where:
- First term \(a = 1\)
- Last term \(l = 49\)
- Common difference \(d = 2\)

The number of terms \(n\) in this sequence can be calculated as:
\[ n = \frac{l - a}{d} + 1 = \frac{49 - 1}{2} + 1 = 25 \]

### Step 5: Calculate the total number of ordered pairs
The total number of ordered pairs \((m,n)\) is given by:
\[ \text{Total pairs} = \text{Number of choices for } m \times \text{Number of choices for } n = 50 \times 25 = 1250 \]

### Final Answer:
Thus, the number of ordered pairs \((m,n)\) such that \(6^m + 9^n\) is a multiple of \(5\) is \(1250\).

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