If the integers m and n are chosen at random between 1 and 100, then the probability that a number of the form `7^(m) + 7^(n)` is divisible by 5, equals :

Options

A. `1//4`
B. `1//7`
C. `1//8`
D. `1//49`

Answer

To solve the problem of finding the probability that the expression \( 7^m + 7^n \) is divisible by 5 when \( m \) and \( n \) are chosen randomly from integers between 1 and 100, we can follow these steps:

### Step 1: Understand the Problem
We need to find the probability that \( 7^m + 7^n \) is divisible by 5. This means we need to analyze the last digit of \( 7^m \) and \( 7^n \) modulo 5.

### Step 2: Find the Pattern of \( 7^k \mod 5 \)
Calculate \( 7^k \mod 5 \) for different values of \( k \):
- \( 7^1 \mod 5 = 2 \)
- \( 7^2 \mod 5 = 4 \)
- \( 7^3 \mod 5 = 3 \)
- \( 7^4 \mod 5 = 1 \)

The pattern repeats every 4 terms:
- \( 7^1 \equiv 2 \)
- \( 7^2 \equiv 4 \)
- \( 7^3 \equiv 3 \)
- \( 7^4 \equiv 1 \)

### Step 3: Determine the Values of \( m \mod 4 \)
From the pattern, we can see:
- If \( m \equiv 1 \mod 4 \), then \( 7^m \equiv 2 \)
- If \( m \equiv 2 \mod 4 \), then \( 7^m \equiv 4 \)
- If \( m \equiv 3 \mod 4 \), then \( 7^m \equiv 3 \)
- If \( m \equiv 0 \mod 4 \), then \( 7^m \equiv 1 \)

### Step 4: Find Combinations for Divisibility by 5
We need to find pairs \( (m, n) \) such that \( 7^m + 7^n \equiv 0 \mod 5 \):
- \( 2 + 3 \equiv 0 \mod 5 \) (i.e., \( m \equiv 1, n \equiv 3 \) or vice versa)
- \( 4 + 1 \equiv 0 \mod 5 \) (i.e., \( m \equiv 2, n \equiv 0 \) or vice versa)

### Step 5: Count the Favorable Outcomes
Each residue class \( \mod 4 \) has 25 integers between 1 and 100:
- \( m \equiv 1 \mod 4 \): 25 choices
- \( m \equiv 2 \mod 4 \): 25 choices
- \( m \equiv 3 \mod 4 \): 25 choices
- \( m \equiv 0 \mod 4 \): 25 choices

Now, we can count the favorable pairs:
1. For \( (m \equiv 1, n \equiv 3) \): \( 25 \times 25 = 625 \)
2. For \( (m \equiv 3, n \equiv 1) \): \( 25 \times 25 = 625 \)
3. For \( (m \equiv 2, n \equiv 0) \): \( 25 \times 25 = 625 \)
4. For \( (m \equiv 0, n \equiv 2) \): \( 25 \times 25 = 625 \)

Total favorable outcomes = \( 625 + 625 + 625 + 625 = 2500 \).

### Step 6: Calculate Total Outcomes
The total number of ways to choose \( m \) and \( n \) is \( 100 \times 100 = 10000 \).

### Step 7: Calculate the Probability
The probability \( P \) that \( 7^m + 7^n \) is divisible by 5 is given by:
\[ P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{2500}{10000} = \frac{1}{4}. \]

### Conclusion
Thus, the probability that \( 7^m + 7^n \) is divisible by 5 is \( \frac{1}{4} \).