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! :)