Modern Algebra

Reply #5

hypersoniq — Fri, 12 Dec 2025 02:57:05 GMT

Mod 10 works because the digits in the game are 0 to 9. You need to compensate for the range of 1 to 49.

The general idea is

(mod(sum - 1, N) + 1)

Say you were summing the last 2 draws and the last digit was 49 both times...

49 + 49 = 98, this is your sum

Subtracting 1 leaves 97 as your sum-1 from above.

N is the highest ball in the game, in your case, 49.

So (mod(97, 49) + 1) would come out to be a 1, which is in the acceptable range of the balls to select. Mod(98,

Reply #4

hypersoniq — Fri, 12 Dec 2025 02:20:22 GMT

Also

T(v) = (v + 3, v + 3, v + 3) mod 10

gives the flip number...

I also recall hearing of a shadow number, but I have not seen a breakdown of that one.

Reply #3

hypersoniq — Wed, 10 Dec 2025 19:59:37 GMT

Cool general workout method!

T(v) = (v + 5, v + 5, v + 5) mod 10

Gives the mirror number.

Reply #2

lakerben — Wed, 10 Dec 2025 17:32:21 GMT

Interesting

Reply #1

Elizabeth03 — Fri, 05 Dec 2025 11:43:33 GMT

How would this work in an Atlantic 49 game? Thank you

Modern Algebra

tokecap — Wed, 03 Dec 2025 10:26:04 GMT

Simple example of *abstract algebra with modular groups* to *Pick 3*, using the group *Z * (integers modulo 10), since the lottery digits range from 0 to 9.

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*1. Defining the group: Z *

Z = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, with *addition modulo 10*

This group is *cyclic* and can be generated by *1 or 3*, for example:

- Generator 1: 1, 2, 3, ..., 9, 0

- Generator 3: 3, 6, 9, 2, 5, 8, 1, 4, 7, 0

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*2. Example of a drawn number: 7-3-9*

Let's represent it as a *v... [ More ]