If the integers m and n are chosen at random between 1 and 100, then atmost distinct numbers of the form `7^m + 7^n` is divisible by 5 equals to :

Options

A. 1250
B. 10000
C. 2500
D. none of these

Answer

To solve the problem of finding the maximum distinct numbers of the form \(7^m + 7^n\) that are divisible by 5, we can follow these steps:

### Step 1: Determine the pattern of \(7^k \mod 5\)
First, we need to evaluate \(7^k \mod 5\) for various values of \(k\):
- \(7^1 \mod 5 = 2\)
- \(7^2 \mod 5 = 4\)
- \(7^3 \mod 5 = 3\)
- \(7^4 \mod 5 = 1\)

Notice that the powers of 7 modulo 5 repeat every 4 terms. The sequence is:
- \(7^1 \equiv 2 \mod 5\)
- \(7^2 \equiv 4 \mod 5\)
- \(7^3 \equiv 3 \mod 5\)
- \(7^4 \equiv 1 \mod 5\)

### Step 2: Identify the residues for \(m\) and \(n\)
Since \(m\) and \(n\) can take any integer value from 1 to 100, we can find the residues of \(m\) and \(n\) modulo 4. The residues can be \(1, 2, 3,\) or \(0\) (where \(0\) corresponds to \(7^4\)).

### Step 3: Calculate \(7^m + 7^n \mod 5\)
Now, we can find the possible values of \(7^m + 7^n \mod 5\) based on the residues:
- If \(m \equiv 1\) and \(n \equiv 1\): \(2 + 2 \equiv 4 \mod 5\)
- If \(m \equiv 1\) and \(n \equiv 2\): \(2 + 4 \equiv 1 \mod 5\)
- If \(m \equiv 1\) and \(n \equiv 3\): \(2 + 3 \equiv 0 \mod 5\)
- If \(m \equiv 1\) and \(n \equiv 0\): \(2 + 1 \equiv 3 \mod 5\)
- If \(m \equiv 2\) and \(n \equiv 2\): \(4 + 4 \equiv 3 \mod 5\)
- If \(m \equiv 2\) and \(n \equiv 3\): \(4 + 3 \equiv 2 \mod 5\)
- If \(m \equiv 2\) and \(n \equiv 0\): \(4 + 1 \equiv 0 \mod 5\)
- If \(m \equiv 3\) and \(n \equiv 3\): \(3 + 3 \equiv 1 \mod 5\)
- If \(m \equiv 3\) and \(n \equiv 0\): \(3 + 1 \equiv 4 \mod 5\)
- If \(m \equiv 0\) and \(n \equiv 0\): \(1 + 1 \equiv 2 \mod 5\)

### Step 4: List the distinct sums
From the calculations above, we can summarize the distinct sums we found:
- \(0\)
- \(1\)
- \(2\)
- \(3\)
- \(4\)

### Step 5: Count the distinct sums
The distinct sums of the form \(7^m + 7^n\) that are divisible by 5 are \(0, 1, 2, 3, 4\). Thus, there are a total of 5 distinct sums.

### Conclusion
The maximum number of distinct integers of the form \(7^m + 7^n\) that are divisible by 5 is **5**.

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