Python Forum

I need help creating a boolean function that will return True if the input has an odd number of digits. No luck so far. Anyone know how to do it, without using a while loop, or advanced python functions?

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I don't really have an idea of what to do. I am currently testing numbers with the floor division and modulus functions. I have found some similar ideas online, but all contained the while loop which I do not have knowledge of. From what I have been trying out, I think the solution to this function has something to do with the floor division function, but I haven't been able to find solid proof.

hint to start: assuming function takes an int as argument, convert it to str and find the length of that str.

Thanks very much for that hint! Helped me a lot.

def odd_number_digits(n):
    if (len(str(n)) % 2 != 0
        return True

This is what I have so far, however, I keep getting a syntax error message referring to my return line. Is there a visible syntax issue?

check the number of closing brackets on the previous line or just remove the most-left opening bracket - it's redundant

Also the function may be simplified to

def odd_number_digits(n):
    return len(str(n)) % 2 != 0
    # return bool(len(str(n)) % 2)

if __name__ == '__main__':
    print(odd_number_digits(123))
    print(odd_number_digits(12))