If two distinct numbers m and n are chosen at random form the set {1, 2, 3, …, 100}, then find the probability that `2^(m) + 2^(n) + 1` is divisible by 3.

Explanation

To find the probability that \(2^m + 2^n + 1\) is divisible by 3 when \(m\) and \(n\) are distinct numbers chosen from the set \(\{1, 2, 3, \ldots, 100\}\), we can follow these steps:

### Step 1: Understanding the Expression
We need to analyze the expression \(2^m + 2^n + 1\) modulo 3.

### Step 2: Calculate \(2^k \mod 3\)
First, we find the pattern of \(2^k \mod 3\):
- \(2^1 \equiv 2 \mod 3\)
- \(2^2 \equiv 1 \mod 3\)
- \(2^3 \equiv 2 \mod 3\)
- \(2^4 \equiv 1 \mod 3\)

This shows that:
- If \(k\) is odd, \(2^k \equiv 2 \mod 3\)
- If \(k\) is even, \(2^k \equiv 1 \mod 3\)

### Step 3: Analyze the Cases
Now we analyze the cases for \(m\) and \(n\):
1. **Both \(m\) and \(n\) are odd**:
\[ 2^m + 2^n + 1 \equiv 2 + 2 + 1 \equiv 5 \equiv 2 \mod 3 \quad (\text{not divisible by 3}) \]

2. **Both \(m\) and \(n\) are even**:
\[ 2^m + 2^n + 1 \equiv 1 + 1 + 1 \equiv 3 \equiv 0 \mod 3 \quad (\text{divisible by 3}) \]

3. **One is odd and the other is even**:
\[ 2^m + 2^n + 1 \equiv 2 + 1 + 1 \equiv 4 \equiv 1 \mod 3 \quad (\text{not divisible by 3}) \]

From this analysis, we conclude that \(2^m + 2^n + 1\) is divisible by 3 only when both \(m\) and \(n\) are even.

### Step 4: Count the Even Numbers
In the set \(\{1, 2, 3, \ldots, 100\}\), the even numbers are:
\[ 2, 4, 6, \ldots, 100 \]
There are 50 even numbers in total.

### Step 5: Calculate the Total Outcomes
The total ways to choose 2 distinct numbers from 100 is given by:
\[ \binom{100}{2} = \frac{100 \times 99}{2} = 4950 \]

### Step 6: Calculate the Favorable Outcomes
The number of ways to choose 2 distinct even numbers from the 50 even numbers is:
\[ \binom{50}{2} = \frac{50 \times 49}{2} = 1225 \]

### Step 7: Calculate the Probability
The probability \(P\) that \(2^m + 2^n + 1\) is divisible by 3 is given by:
\[ P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{1225}{4950} \]

### Step 8: Simplify the Probability
Simplifying \(\frac{1225}{4950}\):
\[ \frac{1225 \div 245}{4950 \div 245} = \frac{5}{20} = \frac{1}{4} \]

### Final Answer
Thus, the probability that \(2^m + 2^n + 1\) is divisible by 3 is:
\[ \frac{1}{4} \]