Fill in the blanks to define a function that prints "Yes", if its parameter is an even number, and "No" otherwise.

Question

Fill in the blanks to define a function that prints "Yes", if its parameter is an even number, and "No" otherwise. please can someone help me with the code?

even(x):
if x%2 == 0:

("Yes")

print("No") my input are def, print else it will marked it green but still not correct

Answer 1

You just forgot to place a colon after your else statement.
I did the same mistake.

def even(x):
if x%2 == 0:
print("Yes")
else:
print("No")

#This should work, By giving "No output"

Answer 2

<form>
<input type="text" name="color"

list

="colors" />
<datalist id="

colors

">
<option value="Red">
<option value="Blue">
<option value="Green">
</

datalist

>
</form>

Answer 3

input is type string , if compare x%2 you need to convert to int
try this
def even(value):
if value % 2== 0:
print("\n" , value, "is Even.")
else:
print("\n" , value, "is Odd.")

x = input("Enter value : ")
print(int(x))
even(int(x))

Answer 4

def
print
else:

Answer 5

def even(x):
if x%2 == 0:
print("Yes")
else:
print("No")

Answer 6

def
print
else:

Answer 7

def even(x):
if x%2 == 0:
print("Numero Par ")
else:
print("Numero impar")

x = int(input("Deme un numero: "))
even(x)

Funciona

Answer 8

def even(x):
if x%2 == 0:
print ("Yes")
else:
print("No")

Answer 9

jtrjy-

Answer 10

Fill in the blanks to associate the input with the datalist
<form>
<input type="text" name="color"

="colors" />
<datalist id="

">
<option value="Red">
<option value="Blue">
<option value="Green">
</

>
</form>
i need unswer pleas