def even_odd_decimal(num): if num%1!=0: return num "is decimal" elif num%2==0: return num "is even" else: return num "is odd"

Question

def even_odd_decimal(num): if num%1!=0: return num "is decimal" elif num%2==0: return num "is even" else: return num "is odd"
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Solution 1

The function even_odd_decimal(num) is intended to check if a number is decimal, even or odd. However, there are some syntax errors in the function. Here is the corrected version:

def even_odd_decimal(num):
    if num % 1 != 0:
        return str(num) + " is decimal"
    elif num % 2 == 
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