Question

Write a program that generates a random number in the range of 1 through 100, and asks the user to guess what the number is. When the number is generated by the computer, don’t display the number to the user, but only display if the number generated is odd or even.

If the user’s guess is higher than the random number, the program should display “Too high, try again.” If the user’s guess is lower than the random number, the program should display “Too low, try again.” Keep asking the user till the correct number is guessed. If the user enters 0 for the guess, the game terminates.

When the user successfully guesses the number, the application should congratulate the user and then display how many tries it took the user to guess the number. Moreover, display all the odd or even numbers starting at 1, up to that number. (Display all odd numbers if the number itself was odd, or all even numbers if the number was even). For example, assume that 16 was the generated number. The final display should be:

You took x tries to guess the number.

16 is an even number.

All even numbers up to 16 are: 1, 2, 4, 6, 8, 10, 12, 14

To generate the random number, you need to import the random library   (import random) and use the randint function ( random.randint(1, 100) )

Answer 1

#Python code

import random

Number = random.randint(1, 100)
if Number %2 == 0:
    print("It is an even Number")
else:
    print("It is an odd Number")

tries = 0

while True:
    tries+=1
    guess = int(input('Enter a Guess (0 to Exit): '))
    if guess == 0:
        break
    elif guess == Number:
        print("You took %d tries to guess the number." %tries)
        break
    elif guess < Number:
        print("Too low, try again.")

    else:
        print("Too high, try again.")

if Number%2 == 0:
    print("All even numbers up to %d are: "%Number ,end = "")
    for i in range(2,Number,2):
        print(i,end=", ")

else:
    print("All odd numbers up to %d are: "%Number ,end = "")
    for i in range(1,Number,2):
        print(i,end=", ")

#screenshot