Interesting number theory question

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Homework Help Overview

The discussion revolves around determining the natural numbers n for which (10^n)-1 is divisible by 73, a problem situated within number theory.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore specific values of n, such as powers of two and multiples of 8, while questioning the validity of their findings and the implications of Fermat's Little Theorem.

Discussion Status

There is an ongoing exploration of various values of n, with some participants suggesting specific numbers and others reflecting on the implications of their calculations. The discussion includes considerations of both larger and smaller values of n, but lacks a definitive consensus on the complete set of solutions.

Contextual Notes

Participants are navigating through the implications of modular arithmetic and the properties of prime numbers, with references to specific mathematical theorems and their applications in this context.

lugita15
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Homework Statement


For what natural numbers n is (10^n)-1 divisible by 73?


Homework Equations


N/A


The Attempt at a Solution


I have already found that it holds when n is a power of two greater than 8. (That means when n is great than 8, not the eigth power of 2)
What other natural numbers n satisfy the above condition?
 
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How about 24,40,48 etc?
 
Not sure how I got there but I think I have the right answer... thought it could be trivialized by Fermat's Little Theorem.

[tex]10^{\phi(73)} \equiv 1 (mod 73)[/tex]
since 73 is prime
[tex]\phi(73) = 72[/tex]

So n=72 works... But Dick commented about numbers less then 72 so I figured it couldn't be that... I took the gcd(72,24) got 24, gcd(72,40) got 8, gcd(72,48) got 24. Then I took the gcd(24,8) and got 8, so it's every multiple of 8 -- and that worked but not really not sure maybe someone else can make sense of it...
 
Once you have 10^8(mod 73)=1 the rest is pretty much a foregone conclusion.
 

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