4 ms·
Ask HN: Mathematicians, Why 9 cancels out in an addition of digits?
I was playing with additions and it seems that 9 cancels itself out (Sorry, nobody thought me this in elementary algebra, or I forgot) when continuously adding all the digits in a number in order to reach one final one-digit number.
Example:
92. 9+2=11 -> 1+1=2. So just remove the 9 and you have 2
989. 17+9 -> 8 + 9 => 1 + 7 = 8. So just leave out the 9s.
94968594892 -> 9+4+9+6+8+5+9+4+8+9+2 = 73 -> 7+3 = 10 -> 1+0 =1
Remove the 9s
4+6+8+5+4+8+2 = 37 -> 3+7 = 1
So why do the smallest one digit number (0) and the largest one digit number (9) both cancel out in such additions?
- meursault334 11y agoBecause adding 9 to any other digit will produce 1 and the other digit - 1. Add these together and you have the original other digit. Because your rule necessitates repeating until you only have 1 digit it doesn't matter when this resolves itself. For example: 9 6=> 1 5 => 6 Put another way, a nine will bump up the 10s place by one and lower the ones place by 1. These cancel out.
- rokhayakebe 11y agoIs there a simple explanation as to why 9 and 0 share this same property?
- meursault334 11y ago0 + other digit = other digit 9 + other digit = (other digit -1) + 1 = other digit I don't know if there is a good way to quantify this other than it is a basic property of base 10 when you are basically looping around the basis. The other post illuminates on this in that it would be E for hex or 5 for base 6.
- dynan 11y agoThink on the modulo or the set of characters that you use for the representation. If you use base 9, it would be 8, for base 6 would be 5 and so on. Or look at it as placed into a revolution circle, or exponentials sequence. In quantifying the multiplicity 9 would be the previous value for 1, not 0