Prove that x^100 = 1 for all x in U(1000)

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Homework Help Overview

The problem involves proving that \( x^{100} = 1 \) for all \( x \) in \( U(1000) \), where \( U(1000) \) is defined as the set of integers less than 1000 that are relatively prime to 1000.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the definition of \( U(1000) \) and question whether certain numbers, such as 1001, should be included. There is confusion about the implications of the problem statement and the application of Euler's Theorem. Some participants attempt to verify the equation with specific examples like \( 3^{100} \) and \( 7^{100} \).

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and clarifying the definition of \( U(1000) \). Some guidance has been offered regarding the use of Euler's Theorem, but no consensus has been reached on the correct approach or assumptions.

Contextual Notes

There is a noted confusion regarding the completeness of the problem statement and the specific properties of numbers in \( U(1000) \). Participants are also grappling with the implications of modular arithmetic in relation to the problem.

nowimpsbball
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Homework Statement



Prove that x^100 = 1 for all x in U(1000)

Homework Equations





The Attempt at a Solution



U(1000) is all numbers relatively prime to 1000. And that is as far as I have gotten.

Thanks
 
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Are you sure you haven't missed a part of the problem...?

x=1001 is relatively prime to 1000, but 1001^100 is not equal to one.
 
Huh? Well U(1000) is all numbers less than a 1000 and relatively prime to 1000, that changes things, at least with your counter example...forgot to mention that
 
What about 3, 7, etc? These are all relatively prime to 1000 (and also less than 1000), yet neither 3^100 nor 7^100 are equal to 1. You are most likely missing some part of the problem.

Edit: You probably meant this - "Prove that x^100 = 1 mod(1000) for all x in U(1000)." Try using Euler's Theorem.
 
Last edited:
Well that is the problem how it is stated...but the idiot I am keeps forgetting to mention all the details of U(n)...it is the group of units modulo n (that is the set of integers less than n and relatively prime to n under multiplication modulo n). But what you say is what I want.
I for example take 3^100=1mod(1000) , 1000 should divide (3^100 - 1) but it does not, you end up with a decimal...so I can't even see what the problem looks like when you plug in a number to x
 
nowimpsbball said:
Well that is the problem how it is stated...but the idiot I am keeps forgetting to mention all the details of U(n)...it is the group of units modulo n (that is the set of integers less than n and relatively prime to n under multiplication modulo n). But what you say is what I want.
I for example take 3^100=1mod(1000) , 1000 should divide (3^100 - 1) but it does not, you end up with a decimal...so I can't even see what the problem looks like when you plug in a number to x

It does divide 3^100 - 1. Not sure what you mean by getting a decimal. If you have Windows, try the built-in calculator and calculate 3^100 mod(1000). You'll get 1, which verifies the equation 3^100=1mod(1000). If you calculate 3^100 - 1 mod(1000), you will get 0, which means that 1000 evenly divides 3^100 - 1.
 

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