Complete Set of Representatives Problems

• November 1st 2010, 03:51 PM
drichi49
Complete Set of Representatives Problems
For which exponents k is {1^k, 2^k, 3^k, 4^k, 5^k, 6^k, 7^k, 8^k, 9^k, 10^k, 11^k} a complete set of representatives modulo 11?

At this point in our course we have covered Induction, Euclid's Algorithm, Unique Factorization, Congruence, Congruence Classes, and Rings/Fields.

I have tried solving this problem several ways. I know that the set is a complete representatives if the set consists of 11 integers (which it does) and no integer in the set is congruent to any other integer in the set. So, i have set a^k=(congruent)b^k (mod 11) where a,b E Set with a>b, and then tried to use this to figure for which values of k this can be true (and then by finding that all other k make the set a complete set of representatives) but I cannot seem to find a way to solve for k.

Any help would be very much appriciated! Thanks!
• November 1st 2010, 04:34 PM
Drexel28
Quote:

Originally Posted by drichi49
For which exponents k is {1^k, 2^k, 3^k, 4^k, 5^k, 6^k, 7^k, 8^k, 9^k, 10^k, 11^k} a complete set of representatives modulo 11?

At this point in our course we have covered Induction, Euclid's Algorithm, Unique Factorization, Congruence, Congruence Classes, and Rings/Fields.

I have tried solving this problem several ways. I know that the set is a complete representatives if the set consists of 11 integers (which it does) and no integer in the set is congruent to any other integer in the set. So, i have set a^k=(congruent)b^k (mod 11) where a,b E Set with a>b, and then tried to use this to figure for which values of k this can be true (and then by finding that all other k make the set a complete set of representatives) but I cannot seem to find a way to solve for k.

Any help would be very much appriciated! Thanks!

I am not sure what the question is asking. Are you asking for which $k\in\mathbb{N}$ is $f:\mathbb{Z}_{11}\to\mathbb{Z}_{11}:z\mapsto z^k$ a bijection?
• November 1st 2010, 04:38 PM
drichi49
Yes, I believe so..
I think that is the case yes. The question is what restrictions exist on k such that {1^k,2^k,...,11^k} a complete set of representatives modulo 11.
• November 1st 2010, 04:39 PM
drichi49
Could you quickly describe what you mean that statement, I am not sure exactly what a map is and what bijection means.
• November 1st 2010, 04:47 PM
Drexel28
Quote:

Originally Posted by drichi49
Could you quickly describe what you mean that statement, I am not sure exactly what a map is and what bijection means.

Please, take note that I am not very knowledgeable about number theory and so things I am saying could be gibberish....

A map is a function and bijection is a one-to-one onto mapping.
• November 1st 2010, 04:55 PM
drichi49
Yes, that is exactly what I am asking then!!!
• November 1st 2010, 05:58 PM
roninpro
Did you just try some values of $k$ to see what happens?