+0

# Euler's Totient Function Problems

0
1
4
+250

For a positive integer $n$, $\phi(n)$ denotes the number of positive integers less than or equal to $n$ that are relatively prime to $n$.
What is $\phi(1200)$?

What is $\phi(181)$?

What is $\phi(212)$?

Jul 26, 2024

#1
+1659
+1

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! :)

Jul 26, 2024
edited by NotThatSmart  Jul 26, 2024

#1
+1659
+1

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! :)

NotThatSmart Jul 26, 2024
edited by NotThatSmart  Jul 26, 2024
#2
+129806
0

Excellent, NTS !!!!

{Without the Totient Function , this would be very tedious....  Euler was a pretty smart guy......!!! }

CPhill  Jul 26, 2024
#3
+1659
+1

"Hey, it's still possible to do it without the function" - famous last words ðŸ˜‚

Thanks, CPhill!

~NTS

NotThatSmart  Jul 26, 2024
edited by NotThatSmart  Jul 26, 2024
#4
+250
+1

Thank You! :)

Jul 29, 2024
edited by BRAINBOLT  Jul 29, 2024