Example on Euler Totient function

Question

I was given in the exam to calculate $\phi(27)$, so I answered like this as I learned:

$\phi(27) = 3^2.3 = (3^2-3^1).(3^1-3^0) = 6-2 = 12$

I got shocked that the answer was supposed to be 18. Can someone explain how we get 18?

Answer 1

For all $n\ge 2$ you have $$ \phi(n)=n\prod_{p\mid n}\left(1-\frac{1}{p}\right), $$ hence $\phi(3^3)=2\cdot 3^2$.

Answer 2

The formula you learned takes the prime factorization, and for that it is necessary for all prime factors to only appear once, with the exponent as big as possible! So:

$\phi(27) = \phi(3^{3}) = (3^{3}-3^{2}) = 18$

Answer 3

TYou have to count how many numbers from 1 .. 27 are not divisible by $3$, so $3=1.3, 6=2.3, 9=3.3, 12=4.3,\ldots, \frac{27}{3}.3$ dont count, so there remain $27 - \frac{27}{3}$ numbers coprime to $3$. But it is better to remember the formula $\phi(p^k) = p^k-p^{k-1}$ (but I prefer $=p^{k-1}(p-1)$).