Euler's Totient Function Problems

Euler's Totient Function! The savior!

Let's look at 1200 first.

First, let me explain the process. 

We must prime factorize every number first. For 1200, we have

$1200 = 12 * 5^2 * 2^2 = 2^4 * 3 * 5^2$

Now, to summarize what we do, we essentially

$\frac{\text{number we start with}}{\text{product of the prime factors}} \cdot \text{product of every prime factor -1}$

So for 1200, we have $\phi{1200} = 1200 / ( 2 * 3 * 5) * (2-1)(3-1) (5-1) = 40 (1)(2)(4) = 320$

Meaning the answer is 320. 

Now, let's do the same for the other two numbers. 

First off, 181. It's a prime number, so the only prime factor is 181. Thus, we have

$\phi 181 = \frac{181}{181}\cdot (181-1) = 180$. So 180 is the answer. 

Now, let's also do 212. We have that $212 = 2^2*53$

Thus, using Euler's Totient Function, we have

$\phi212 = \frac{212}{2*53} \cdot (2-1)(53-1) = 2*52 = 104$ so 104 is the answer. 

Thus, the 3 answers are

$\phi1200 = 320\\ \phi 181 = 180\\ \phi 212 = 104$

Thanks! :)

Thank You! :)