this post was submitted on 04 Nov 2024
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If all of the digits summed recursively reduce to a 9, then the number is divisible by 9 and also by 3.
If the difference between the sums of alternating sets digits in a number is divisible by 11, then the number itself is divisible by 11.
That’s all I can remember, but yay for math right?
Well, on the side of easy ones there is "if the last digit is divisible by 2, whole number is divisible by 2". Also works for 5. And if you take last 2 digits, it works for 4. And the legendary "if it ends with 0, it's divisible by 10".
Its never divisible by zero, and its always divisible by one
Never say never. 1/0 = 0
Interesting read. Thank you.