Else if statement in Python

Question

This is what the program wants me to output

Give a number: 10 Give another number: 11 One of the numbers is even.

Give a number: 12 Give another number: 20 Both numbers are even.

Give a number: 15 Give another number: 21 Both numbers are odd.

Here is what I tried to do

number1 = int(input("Give a number:"))
number2 = int(input("Give another number:"))
if (number1 % 2 == 0) or (number2 % 2 ==0):
        print("One of the numbers is even.")
elif (number1 % 2 != 0) or (number2 % 2 !=0):
            print("One of the numbers is odd)

But the program said I have a syntax error *Edit: I tried programming this but it doesn't give out the output I want

Please provide the full error message, it will likely tell you exactly where the problem is — Mark Harwood
– Mark Harwood, Commented Nov 11, 2019 at 16:34

Uber · Accepted Answer · 2019-11-11 16:34:34Z

2

You forgot one " at the last print ;)

number1 = int(input("Give a number:"))
number2 = int(input("Give another number:"))
if (number1 % 2 == 0) or (number2 % 2 ==0):
        print("One of the numbers is even.")
elif (number1 % 2 != 0) or (number2 % 2 !=0):
            print("One of the numbers is odd")

input: 1 & 10 output: One of the numbers is even.

answered Nov 11, 2019 at 16:34

Uber

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Comments

timosmd · Accepted Answer · 2019-11-11 16:35:37Z

1

There is a " missing in your last print statement.

It should be: print("One of the numbers is odd")

answered Nov 11, 2019 at 16:35

timosmd

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