Calculate $\phi(36)$, where $\phi$ is the Euler Totient function. Use this to calculate $13788 \pmod {36}$. I am wondering if anyone can help me. I am trying to figure out how to

Calculate $\phi(36)$, where $\phi$ is the Euler Totient function. Use this to calculate $13788 \pmod {36}$.

I have an exam coming up an this will be one style of question. Can anyone please walk me through how it is done?
Thanks to SchrodingersCat I now know first part is $12$.
The second part should be along the lines of below but I do not understand how this was arrived at.
\begin{align}
13788 \pmod {36} &= 13788 \pmod {\phi(36)} \pmod {36} \\ &= 13788 \pmod {12} \pmod {36} \\ &= 138 \pmod {36} \\ &= ((132)2)2 \pmod {36} \\ &= (252)2 \pmod {36} \\ &= 132 \pmod {36} \\ &= 25
\end{align}
Can anyone show me why it is $25$ and how do I get it?
 A: $36=4\times9$
Since $4$ and $9$ are relatively prime, we have
$\phi(36)=\phi(4)\phi(9)$
Using the identity that for any prime $p$,  $\phi(p^n)=p^n-p^{n-1}$
$\phi(4)\phi(9)=\phi(2^2)\phi(3^2)=(2^2-2).(3^2-3)=12$
As for the second part of the question, then, 13788 (mod 12) (mod 36) $\ne
138(mod 36)$ and hence your question is wrong.
There seems to be typo error and this part of the question can be modified to $13^{788}(mod\ 36) = 13^{788(mod \phi(36))}(mod\ 36) = 13^{788(mod 12)}(mod\ 36)=13^8(mod\  36)=25$
A: $$36=2^2\cdot 3^2$$
So $$\phi(36)=36\cdot \left(1-\frac{1}{2}\right)\left( 1-\frac{1}{3}\right)=12$$
If you are unaware of this formula, then check this link and also the example there.
For the second part, observe that $13788=2^2\cdot 3^2\cdot 383$ 
So $36$ divides $13788$.
That is, $$13788 \equiv 0 \pmod {36}$$
A: Remember that:
$\phi(p)=p-1$ 
$\phi(p^n) = p^n -p^{n-1}$
for every prime number $p$. 
And $\phi(ab)=\phi(a)\phi(b)$ whenever $a$ and $b$ are relatively prime.
Thus $\phi(36) = \phi(9)\phi(4)=\phi(3^2)\phi(2^2)=(9-3)(4-2)=12$.
We are lucky since
$13788 = 383\cdot 36$
and so
$13788 = 383\cdot 36 \equiv 0 \pmod{36}$.
If you want to compute $a^{13788}$ modulo $36$ when $a$ and $36$ are relatively prime, you can use Little Fermat Theorem which says that
$a^{\phi(36)} \equiv 1 \pmod{36}$.
You get
$a^{13788} = (a^{12})^{1149} \equiv 1^{1149} \equiv 1 \pmod{36}$.
