# Eternity Async Tutoring

**URL:** https://discuss.criticalfallibilism.com/t/eternity-async-tutoring/1932
**Category:** Tutoring
**Created:** [April 17, 2024, 11:22pm UTC](https://discuss.criticalfallibilism.com/t/eternity-async-tutoring/1932 "2024-04-17T23:22:52Z")
**Posts on this page:** 1
**Showing post:** 443

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### Author: ![Eternity](https://discuss.criticalfallibilism.com/user_avatar/discuss.criticalfallibilism.com/eternity/32/5930_2.png) [@Eternity](https://discuss.criticalfallibilism.com/u/Eternity)
#### Post date: [July 22, 2024, 12:04am UTC](https://discuss.criticalfallibilism.com/t/eternity-async-tutoring/1932/443 "2024-07-22T00:04:20Z")

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> [@Elliot](#):
>
> Does your solution rely on negative numbers or could you just add 2 so it deals with positives so you don’t have to worry about mod/remainder weirdness? In other words, would the below work? You tell me.
> 
> (x-y+2) remainder 2

It does rely on negative numbers. Let’s see  
(0-0+2) remainder 2 = 0  
(1-0+2) remainder 2 = 1 + 2 remainder 2 = 3 remainder 2 = 1  
(0-1+2) remainder 2 = -1 + 2 remainder 2 = 1 remainder 2 = 1  
(1-1+2) remainder 2 = 0 + 2 remainder 2 = 2 remainder 2 = 0

Hmm. Yeah I could’ve added two and not dealt with negative numbers.

> [@Elliot](#):
>
> take x+y, x+y+1 or x\*y as one thing and then add, subtract or multiply it with a second thing. The second thing could be using remainder 2 on any of the first things. Doing this, you should either be able to find a solution or find something that seems useful which has some advantage over your starting place which could reach a solution using further operations.

or(x,y) = (x+y+xy) remainder 2  
or(0,0) = (0+0+0) remainder 2 = 0  
or(1,0) = (1+0+0) remainder 2 = 1  
or(0,1) = (0+1+0) remainder 2 = 1  
or(1,1) = (1+1+1) remainder 2 = 3 remainder 2 = 1

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