Analyzing the Pseudocode.
Initially, \( x = 0 \), \( flag = 0 \), and \( sum = 0 \).
First, \( x = 0 \) is pushed onto the stack.
The loop starts with \( flag = 0 \), and the value of \( x \) is incremented by 1 on each iteration until \( x = 8 \). So the stack contains: \( [0, 1, 2, 3, 4, 5, 6, 7, 8] \).
When \( x = 8 \), \( flag \) is set to 1.
Now, \( flag = 1 \), and we start popping values from the stack.
First, \( x = 8 \) is popped from the stack.
Since \( x = 8 \) is even, we don't pop again and add \( x \) to the sum. So, \( sum = 0 + 8 = 8 \).
Next, \( x = 7 \) is popped from the stack.
Since \( x = 7 \) is odd, we pop the next value, \( x = 6 \), and add \( x \) to the sum. So, \( sum = 8 + 6 = 14 \).
Next, \( x = 5 \) is popped from the stack.
Since \( x = 5 \) is odd, we pop the next value, \( x = 4 \), and add \( x \) to the sum. So, \( sum = 14 + 4 = 18 \).
Next, \( x = 3 \) is popped from the stack.
Since \( x = 3 \) is odd, we pop the next value, \( x = 2 \), and add \( x \) to the sum. So, \( sum = 18 + 2 = 20 \).
Finally, \( x = 1 \) is popped from the stack.
Since \( x = 1 \) is odd, we pop the next value, \( x = 0 \), and add \( x \) to the sum. So, \( sum = 20 + 4 = 24 \).
Step 2: Conclusion.
The final value of \( sum \) after the program executes is \( \boxed{24} \).
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