The number of ordered pairs `(m,n),m,n in{1,2,...,100}` such that `7^m + 7^n` is divisible by 5 is

Options

A. `1//5`
B. `1//7`
C. `1//4`
D. `1//49`

Answer

To solve the problem of finding the number of ordered pairs \((m, n)\) such that \(7^m + 7^n\) is divisible by 5, we can follow these steps:

### Step 1: Determine the units digits of \(7^m\)
The units digits of powers of 7 repeat every 4 terms. We can calculate them as follows:
- \(7^1 \equiv 7 \mod 10\) (units digit is 7)
- \(7^2 \equiv 49 \mod 10\) (units digit is 9)
- \(7^3 \equiv 343 \mod 10\) (units digit is 3)
- \(7^4 \equiv 2401 \mod 10\) (units digit is 1)

Thus, the units digits of \(7^m\) cycle through \(7, 9, 3, 1\).

### Step 2: Find the conditions for divisibility by 5
For \(7^m + 7^n\) to be divisible by 5, we need to consider the possible units digits:
- \(7 \equiv 2 \mod 5\)
- \(9 \equiv 4 \mod 5\)
- \(3 \equiv 3 \mod 5\)
- \(1 \equiv 1 \mod 5\)

Now, we can summarize the congruences:
- \(7^m \equiv 2\) when \(m \equiv 1 \mod 4\)
- \(7^m \equiv 4\) when \(m \equiv 2 \mod 4\)
- \(7^m \equiv 3\) when \(m \equiv 3 \mod 4\)
- \(7^m \equiv 1\) when \(m \equiv 0 \mod 4\)

### Step 3: Identify pairs that satisfy the divisibility condition
To satisfy \(7^m + 7^n \equiv 0 \mod 5\), we can have the following pairs:
1. \( (2, 3) \) or \( (3, 2) \)
2. \( (4, 1) \) or \( (1, 4) \)

### Step 4: Count the values of \(m\) and \(n\)
For \(m\) and \(n\) ranging from 1 to 100:
- The values of \(m\) congruent to \(1 \mod 4\) are \(1, 5, 9, \ldots, 97\) (25 values).
- The values of \(m\) congruent to \(2 \mod 4\) are \(2, 6, 10, \ldots, 98\) (25 values).
- The values of \(m\) congruent to \(3 \mod 4\) are \(3, 7, 11, \ldots, 99\) (25 values).
- The values of \(m\) congruent to \(0 \mod 4\) are \(4, 8, 12, \ldots, 100\) (25 values).

### Step 5: Calculate the total number of ordered pairs
Using the conditions identified:
1. For pairs \( (m \equiv 1, n \equiv 3) \) or \( (m \equiv 3, n \equiv 1) \):
- \(25 \times 25 + 25 \times 25 = 1250\)
2. For pairs \( (m \equiv 2, n \equiv 0) \) or \( (m \equiv 0, n \equiv 2) \):
- \(25 \times 25 + 25 \times 25 = 1250\)

Adding these gives us:
\[ 1250 + 1250 = 2500 \]

### Step 6: Calculate the total sample space
The total sample space for ordered pairs \((m, n)\) where \(m, n \in \{1, 2, \ldots, 100\}\) is:
\[ 100 \times 100 = 10000 \]

### Step 7: Calculate the probability
The probability that \(7^m + 7^n\) is divisible by 5 is:
\[ \frac{2500}{10000} = \frac{1}{4} \]

### Final Answer
Thus, the number of ordered pairs \((m, n)\) such that \(7^m + 7^n\) is divisible by 5 is \(2500\).

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