hi...so I've attempted this question with what I thought was an elegant solution :)... def don_odd_even(num): if (num % 2) == 0: print('Even') else: print('Odd') ...but reviewing Giles' solution... def odd_even(num): if (num & 1): print('Odd') else: print('Even') ... I'd like to understand how the '&' operator achieved the same functionality as the modulus function used in my attempt. Warm regards, -don

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 bin(a) 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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Don Orazulume

Last answered:

12 Aug 2022

Posted on:

11 Aug 2022

Understanding a better solution Q19 (num & 1)

in Python Programmer Bootcamp / Further exrecises

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Hristina Hristova

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Posted on:

12 Aug 2022

Understanding a better solution Q19 (num & 1)

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