KnifeMissile

I'm more interested in why these mathematical tricks work than their mere existence.
For instance, I've been thinking about this whole "sum of 9's" thing and I think I finally see how it works.

9 is the last digit in the decimal system, so when you add another 9 to it, you necessarily increment the next digit and decrement the current digit. So:

2 * 9 = 18
3 * 9 = 27
4 * 9 = 36

...you get the idea. Because of this digit conservation (one digit place decreases while the other increases by exactly the same amount), they must necessarily sum to 9! (exclamation, not factorial!)

Now, you're probably thinking "but what about when the digits sum to a number greater than nine!" Well, you'll note that they sum to a multiple of nine! This is necessarily true, because of how our digital representation of numbers (yes, numbers do exist beyong the written notation we use to represent them), like so:

let b = 10 (the base)

16384 = b^4 + 6*b^3 + 3*b^2 + 8*b + 4

...now, if you were to multiply 16384 by 9, you will be multiplying each term by 9 (the distributive property of multiplication). But then each term will be made of digits that will sum to 9 because they are only single digit numbers, and we know that those digits will sum to 9, so all these digits put together must sum to some multiple of 9! (exclamation...) We all know what property multiples of 9 have, right? Their digits sum to nine!

Actually, this last part is a bit of circular logic since we're trying to prove that summing the digits of multiples of nine sum to nine but, believe it or not, it's the start of a rigorous proof, because we know that the sum of a number's digits must be less than the number. So, you can use a form of mathematical induction called (if memory serves me) strong induction. Anyway, I hope you all get the point!

Furthermore, because we understand how this works, we can safely say that this property really isn't all that weird. For instance, if you work with hexadecimal numbers, you'll note that the same thing happens with the number F (fifteen). That is, multiply any number with fifteen and then sum the hexadecimal digits of that product. You will get a number that is a multiple of fifteen! Coincidence? No, because 15 is 1 less than the hexadecimal base, 16. Try 7 base 8 (octal). As long as the digit is one less than the base, this trick will work in that base...