Last answered:

12 Aug 2022

Posted on:

11 Aug 2022


Understanding a better solution Q19 (num & 1) I've attempted this question with what I thought was an elegant solution :)...

def don_odd_even(num):
    if (num % 2) == 0:

...but reviewing Giles' solution...

def odd_even(num):
    if (num & 1):

... I'd like to understand how the '&' operator achieved the same functionality as the modulus function used in my attempt.

Warm regards,

1 answers ( 0 marked as helpful)
Posted on:

12 Aug 2022


Hey Don,

Thank you for your question!

Your solution is indeed great! What Giles makes use of are bitwise operators.

If you execute on a new line the following:

42 & 1

you will get 0. If you instead execute the following:

43 & 1

you will get 1. Typing any odd number before & will result in 1, while trying any even number before the & will return 0. Since 1 corresponds to True, while 0 corresponds to False, you will print out 'Odd' for odd numbers and 'Even' for even numbers.

Think of the & operator as binary multiplication. By applying & to num and 1, it is implied that you are multiplying them in binary. In binary, all even numbers end with a 0, while all odd numbers end with a 1. This means that multiplying the former by 1 would return a 0, while multiplying the latter by 1 would return a 1.

You can use the following syntax to convert decimal numbers in binary


where a is any integer. You will quickly convince yourself that indeed all even numbers end with a 0, while all odd numbers end with a 1.

Hope this helps!

Kind regards,
365 Hristina

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