Question

Given positive integer n, write a for loop that outputs the even numbers from n down to 0. If n is odd, start with the next lower even number. Hint: Use an if statement and the % operator to detect if n is odd, decrementing n if so. Enter an integer: 7 Sequence: 64 20 (2) If n is negative, output 0, Hint: Use an if statement to check if n is negative. If so, just set n = 0. Enter an integer: -1 Sequence: 0 Show transcribed image text (1) Given positive integer n, write a for loop that outputs the even numbers from n down to 0. If n is odd, start with the next lower even number. Hint: Use an if statement and the % operator to detect if n is odd, decrementing n if so. Enter an integer: 7 Sequence: 64 20 (2) If n is negative, output 0, Hint: Use an if statement to check if n is negative. If so, just set n = 0. Enter an integer: -1 Sequence: 0

Answer 1

The provided solution involves using an if statement to adjust the initial value of n if it is negative or odd, and then a for loop to print even numbers in descending order from the adjusted n to 0.

Step-by-step explanation:

To address the student's problem of outputting even numbers from n down to 0 using a for loop in a programming context, one can write code that initially checks if n is negative or not, and then proceeds to check if n is even or odd. The loop then iterates in decreasing order, printing even numbers:

if n < 0:
n = 0
elif n % 2 != 0:
n -= 1
for i in range(n, -1, -2):
print(i)

This pseudocode involves an if statement to detect and handle a negative value of n, and another if statement paired with the % operator (modulo operator) to check for oddness and adjust n accordingly. The for loop uses a step of -2 to decrement from the (adjusted) starting point down to 0, ensuring only even numbers are printed.