Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find $n\in \mathbb N$ so that $(\mathbb Z_n, +, \cdot)$ has exactly 4 invertible elements and 5 zero-divisors.

As I couldn't find any theorem that would lead me toward a solution, so I have been trying guessing and checking with no results so far, so I am asking for a push in the right direction.

share|improve this question
1  
Is it exactly 4 invertible elements and exactly 5 zero-divisors? –  lhf Sep 1 '12 at 17:11
    
@lhf yep! I am sure. –  haunted85 Sep 1 '12 at 22:05
    
In this case, $n$ would have to be 9 but $\mathbb Z_9$ has 6 units. –  lhf Sep 2 '12 at 1:07
    
Also, $\phi(n)=4$ iff $n=5, 8, 10, 12$. –  lhf Sep 2 '12 at 1:10
1  
Perhaps we are using a definition whereby zero is not a zero-divisor. –  Gerry Myerson Sep 2 '12 at 6:34

1 Answer 1

HINT: $x$ is a unit in $\mathbb{Z}_n$ if and only if gcd$(n,x) = 1$

EDIT: furthermore: Let $R$ be a finite ring with unity, and $x \in R$: $x$ is a unit in $R$ if and only if $x$ is no zero divisor and $x \neq 0$

share|improve this answer
    
@BenMillwood My bad! –  sxd Sep 1 '12 at 15:27

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.