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## #1 2013-10-09 11:40:13

cooljackiec
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### phi

could someone give me a step by step solution. I figured out
was
which are odd numbers. I am not sure if i could do the same to this and divide it by 3.

Last edited by cooljackiec (2013-10-09 11:40:28)

I see you have graph paper.
You must be plotting something

## #2 2013-10-09 11:59:36

anonimnystefy
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### Re: phi

Hi

The answer is 2*3^99, like you thought it might be. For all primes p it is true that phi(p^n)=p^n-p^(n-1).

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #3 2013-10-09 14:40:44

cooljackiec
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### Re: phi

how would you prove that? by induction?

I see you have graph paper.
You must be plotting something

## #4 2013-10-09 17:02:20

anonimnystefy
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### Re: phi

Not exactly. I'd prove it using your idea. 1/p of the first p^n nunbers are not relatively prime to p^n and the rest is, so phi(p^n)=p^n-(1/p)*p^n=p^n-p^(n-1).

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment