Invers Modulo

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ErikHall

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Nov 11, 2019
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So i am Working on something right now involving some Modulo Operations. If you want to know more, you can ask but its not needed for the Problem. Now this is not a School Problem, its one i "made up" myself, and cannot solve.

So this is the Problem:
Mod.png
I need to solve for x. And i know that x is suppose to be 22. But i dont even know how to start. At least no good way. One way would be to just divide 1077 by 21 and 2 and look if there are any common dividers. But this seems kind of cheap to me.

So yeah, if any of you have a good solution for this, hit me

Thanks for your Help !
 
So for clarification:
I once asked a question on this Forum about a way to encrypt a String of 3 numbers (ranging from 1 to 21) into a single number. And your Members @Cubist created a way to do it. The basic form looks like this: k = n1+n2*t+n3*t². In which k is the resulting Number. n1, n2, n3 are the Numbers, or Coordinates, to encryp. And the is the "divider".
All of this is for a Movie Script, the name is Hexaeder, in which people are trapped inside a giant Maze of Cubical Rooms with Numbers in them. The Numbers inside the Rooms, for Example (1077 / 3369 / 1816) , represent a String of 3 Coordinates, throw which the Room moves. In this case it looks like this (X/Y/Z).
To get the real Coordinates, or the Numbers from n1-n3, you have to divide the Number by t, or 22 in this case. So one Character should find a way to find t, or in the picture x. I think it may be enough to just know 2 Coordinates and then solve for x. But again, i dont know how.
 
[MATH]1077 \pmod{x} \equiv 21\\ 1077 = k x + 21,~\text{ for some $k \in \mathbb{Z}$}\\ k x = 1077-21 = 1056[/MATH]
[MATH]1056 = 2^5 \cdot 3 \cdot 11\\ \text{so $x$ can be any product of some combination of these factors that is 22 or greater}\\ \text{We can list these: 22, 33, 44, 66, 24, 88, 132, 32, 48, 176, 264, 96, 352, 528, 1056}\\ \text{and we can check them in the second modular equation}\\ \text{I find that 528 is the only solution} [/MATH]
Seeing as how you know your solution is 22 I'm probably missing something.
 
Romsek said:
[MATH]1077 \pmod{x} \equiv 21\\ 1077 = k x + 21,~\text{ for some $k \in \mathbb{Z}$}\\ k x = 1077-21 = 1056[/MATH]
[MATH]1056 = 2^5 \cdot 3 \cdot 11\\ \text{so $x$ can be any product of some combination of these factors that is 22 or greater}\\ \text{We can list these: 22, 33, 44, 66, 24, 88, 132, 32, 48, 176, 264, 96, 352, 528, 1056}\\ \text{and we can check them in the second modular equation}\\ \text{I find that 528 is the only solution} [/MATH]
Seeing as how you know your solution is 22 I'm probably missing something.
Click to expand...

The 22 comes from this:
21+4*22+2*22² = 1077
So n1 = 21, n2 =4 and n3 = 2
Now you just need to divide by 22. Like this:
1077 : 22 = 48 / Remainder 21
48 : 22 = 2 / Remainder 4
So this is how you get 21/2/4. They are in the right order one you divide by hand.

Maybe i messed up the Mod in some way. But the idea is that 1077 mod x = 21 is the first division and (1077-21)/x mod x = 2 is the 2. one. But again, maybe i did it wrong.,
 
Romsek said:
[MATH]1077 \pmod{x} \equiv 21\\ 1077 = k x + 21,~\text{ for some $k \in \mathbb{Z}$}\\ k x = 1077-21 = 1056[/MATH]
[MATH]1056 = 2^5 \cdot 3 \cdot 11\\ \text{so $x$ can be any product of some combination of these factors that is 22 or greater}\\ \text{We can list these: 22, 33, 44, 66, 24, 88, 132, 32, 48, 176, 264, 96, 352, 528, 1056}\\ \text{and we can check them in the second modular equation}\\ \text{I find that 528 is the only solution} [/MATH]
Seeing as how you know your solution is 22 I'm probably missing something.
Click to expand...
Ok i f.ed up. ((1077.21/x) mod x = 4 not 2 as 4 is the 2. Number not 2.
 
Ok, in this case the solutions are x=22, x=264

The process is the same.
 
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