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

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?

  • $\begingroup$ do you mean in the ring $\mathbb{Z}_{20}$? $\endgroup$ – KNilesh Sep 10 '19 at 16:12
  • $\begingroup$ 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. $\endgroup$ – JMoravitz Sep 10 '19 at 16:16
  • $\begingroup$ 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. $\endgroup$ – Chris Leary Sep 10 '19 at 16:17
  • $\begingroup$ got it , thanks..there will be 8 such elements and the other 12 are non-invertible $\endgroup$ – Balchandar Reddy Sep 10 '19 at 16:21
  • $\begingroup$ Reference topic : Totient function. That's totient, not quotient. $\endgroup$ – 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 $20$, $\phi(20)=\phi(4)\times\phi(5)=2\times4=8.$


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