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Anonymous Tue 21 Jan 14:54:25 2020 No.11320032 View Reply Original Report

Quoted By: >>11320036 >>11320049 >>11320467

(5+1+x)%17=(8+2+x)%19=(1+3+x)%7=(7+4+x)%13=(1+5+x)%5=(0+6+x)%3=(0+7+x)%11=0

how would i solve this for x by hand? i know i can just type it into wolframalpha but i'm curious to know how it actually works.

Anonymous

Anonymous Tue 21 Jan 2020 14:58:50 No.11320036 Report

Quoted By:

>>11320032
Simply rewrite the form, but I don't want to compromise encryption, as per following thing, rewriting modulus as " + unknown variable * mod " you can crack even RSA.\

What you have proposed here is crystal grid encryption, which can be cracked similary.

Anonymous

Anonymous Tue 21 Jan 2020 15:05:12 No.11320049 Report

Quoted By:

>>11320032
You would rewrite it to basic forms, as 6+x = a * 17 ; 10+x = b * 19, therefore you know there is 16+2x = a * 17 + b * 19

you follow this for every, therefore sum of left side and sum of right side of this cut is equal. you then get polynomial on one side... from that one you can move to reduce

Anonymous

Anonymous Tue 21 Jan 2020 18:38:21 No.11320467 Report

Quoted By:

>>11320032
correct me if im wrong here:

from the last 2 terms we have:
(0+7+x)%11=0
7+x=0
=>x=-7

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