Reverse—Use the inverse.
What is this post about?: I wanted to find inverse functions for digital root and for Factorial (! and dr()) …I will leave it at dr() for now…factorial tomorrow…Really, read this, and it will make you feel smart…see, I really have no life…
Background info:
First, one must have an in depth understanding of both functions. I will start with digital root. Digital root is the sum of all the digits in a number, then the sum of the new number, until the remaining number contains only one digit. ex: dr(32)=3+2=5 dr(19395) = 1+9+3+9+5 = 27 = 9….so dr(19395)=9… This is a very cumbersome process.
Moving away from that for a while….In order to simplify the process, one must understand The modulus fucntion, and using it with 9. I will write modulus as n % x. Modulus is the remainder of the division of 2 numbers. So…5%9=5 easy enough concept.
Now…dr(5) is also == 5… Division by 9 is used to keep it digital…There are 9 digits that will affect the digital root of the number. In more simple terms, there are 9 non-zero numbers. So you may be thinking, that digial root can be expressed as dr(x)=x%9. Well, in most cases that is true, but consider any case where X is 9*10n …That would be very very wrong. infact… 99/9 is 11. THis is due to the division by 9. It will come out to 0. There is no number with a digital root of 0(this would not happen if we divided by 10 ;)) So, if we add one, then we are set for that. It will never come out to one. so far we have: dr(x)=1+(x%9) Well, that seems to be good, except now every number comes out one too many. So, it looks like we also have to subtract one. To conclude: dr(x)=1+mod[(x-1),9]
So what is dr()-1?
Well, it is non-existant, basically. That formula can be re-written in terms of X, But you will end up with the mod-1 which is NULL. The only way the inverse digital root can be done is if one knows what (X-1)/9 is…otherwise, there is no way to reverse mod()
Ok…Tomorrow, I will post about inverse Factorials…and make SubProfile Version 4.0