# Invertibility of integers modulo $20$

Find the Number of elements which are not invertible in the set of integers $$\{0, 1, 2, 3, ..... 19\}$$ modulo $$20$$.

Approach:
I have tried finding the elements which satisfies $$k$$*(element) mod $$20 = 1$$ where $$k>1;$$ the elements which satisfy the equation indeed are the co-primes of $$20.$$
There are $$12$$ such elements. Is this correct?

• do you mean in the ring $\mathbb{Z}_{20}$? – KNilesh Sep 10 '19 at 16:12
• There are $12$ elements which are not coprime with $20$ and these are the noninvertible elements. Meanwhile there are eight elements which are coprime and are the invertible elements. Make sure you understand why and how to count these. – JMoravitz Sep 10 '19 at 16:16
• I found your wording a bit confusing. You are correct that the invertible elements are those prime to 20. There are 8 of them. So, the remaining 12 are not invertible. You can allow $k$ to be 1 since 1 is certainly invertible. The description of your approach, at least at a quick reading, gives the impression you are saying there are 12 elements prime to 20. Always be careful that what you write is what you mean to say. – Chris Leary Sep 10 '19 at 16:17
• got it , thanks..there will be 8 such elements and the other 12 are non-invertible – Balchandar Reddy Sep 10 '19 at 16:21
• Reference topic : Totient function. That's totient, not quotient. – DanielWainfleet Sep 10 '19 at 18:53

The prime factorization of $$20$$ is $$2^2\times5$$, so multiples of $$2$$ and $$5$$ are not invertible modulo $$20$$.
That leaves the following $$8$$ invertible residues modulo $$20$$: $$1, 3, 7, 9, 11, 13, 17$$, and $$19.$$
Accordingly, Euler's totient function of $$20$$, $$\phi(20)=\phi(4)\times\phi(5)=2\times4=8.$$