

Ask HN: Mathematicians, Why 9 cancels out in an addition of digits? - rokhayakebe

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.<p>Example:<p>92. 9+2=11 -&gt; 1+1=2. So just remove the 9 and you have 2<p>989. 17+9 -&gt; 8 + 9 =&gt; 1 + 7 = 8. So just leave out the 9s.<p>94968594892 -&gt; 9+4+9+6+8+5+9+4+8+9+2 = 73 -&gt; 7+3 = 10 -&gt; 1+0 =1<p>Remove the 9s<p>4+6+8+5+4+8+2 = 37 -&gt; 3+7 = 1<p>So why do the smallest one digit number (0) and the largest one digit number (9) both cancel out in such additions?
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meursault334
Because 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.

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rokhayakebe
Is there a simple explanation as to why 9 and 0 share this same property?

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meursault334
0 + 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.

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dynan
Think 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

