Euler's Totient Function

**Maccaman** · March 28th 2009, 06:24 PM

Hi, I was just wondering if someone could confirm that the following is correct? Just a yes or no would suffice.

$\varphi (190) = 72$

Thanks

**Nokio720** · March 28th 2009, 07:16 PM

Yes it looks correct, since

$\varphi(p^k)=(p-1)p^{k-1}$ for a prime p and using the fact that phi is multiplicative i.e
$\varphi(ab)= \varphi(a) * \varphi(b)$

Then

$ \varphi (190) = 72 $

$ \varphi(190)=\varphi(2*19*5)=\varphi(2) * \varphi(5) * \varphi(19) $

$ =1(2^{1-1}) *4(5^{1-1}) *18(19^{1-1}) \ { }= (2^0)*4(5^0)*18(19^0) = 4*18=72 $

Yes your answer is correct. Hope this helps!

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