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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 $\phi(2835)$?

 Jun 20, 2024

Best Answer 

 #1
avatar+1926 
+1

We can use Euler's Totient Function. 

Let's find all the distinct prime factors of 2835 first. 

The prime factors of 2835 are \(3,5,7\)

 

Now, we simplfy do \(2835(1-1/3)(1-1/5)(1-1/7)\)

 

Simplfying this, we have

\(2835 * \frac{2}{3}*\frac{4}{5}*\frac{6}{7} =\frac{2835\cdot \:2\cdot \:4\cdot \:6}{1\cdot \:3\cdot \:5\cdot \:7}=1296\)

 

So 1296 is our answer. 

I'm not sure if I did this correctly...

 

Thanks! :)

 Jun 20, 2024
edited by NotThatSmart  Jun 20, 2024
 #1
avatar+1926 
+1
Best Answer

We can use Euler's Totient Function. 

Let's find all the distinct prime factors of 2835 first. 

The prime factors of 2835 are \(3,5,7\)

 

Now, we simplfy do \(2835(1-1/3)(1-1/5)(1-1/7)\)

 

Simplfying this, we have

\(2835 * \frac{2}{3}*\frac{4}{5}*\frac{6}{7} =\frac{2835\cdot \:2\cdot \:4\cdot \:6}{1\cdot \:3\cdot \:5\cdot \:7}=1296\)

 

So 1296 is our answer. 

I'm not sure if I did this correctly...

 

Thanks! :)

NotThatSmart Jun 20, 2024
edited by NotThatSmart  Jun 20, 2024
 #2
avatar+129895 
0

Good job, NTS!!!!

 

 

cool cool cool

CPhill  Jun 20, 2024
 #3
avatar+1926 
+1

Thanks to you too, CPhill. 

I did not know about Euler's Totient Function until today when we did a problem similar earlier. 

Sure saved me a lot of time on this one! Lol! :)

 

Thanks! :)

 

~NTS

NotThatSmart  Jun 20, 2024
edited by NotThatSmart  Jun 20, 2024

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