Hi Guys,
Any information would be great.
φ$\displaystyle (30)$
Thank you mate,
Looking at this example, can you tell me how they came to use 2 & 3
$\displaystyle \varphi(36)=\varphi\left(2^2 3^2\right)=36\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)=36\cdot\frac{1}{2}\cdot\frac{2} {3}=12.$
So i attempted $\displaystyle \varphi(n)= (10)$
10 = (2 * 5) = 10 * (1 - 1/2)(1 - 1/5)
= 10 * (1/2 * 4/5)
= 4/10 * 10/1
= 4
So there should be 4 prime numbers less than 10 (not including 2 and 5)
1, 3, 7 (only three)
What am i doing wrong?
In my previous post I have been a little 'approximative'... if n is not prime, i.e. is...
$\displaystyle n= p_{1}^{k_{1}}\ p_{2}^{k_{2}}\ ...\ p_{i}^{k_{i}}$ (1)
... then the totiens function is 1 plus the number of primes p and their powers less than n. In Your case You have 'forgotten' 3 x 3=9...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
by the chinese remainder theorem, if p,q are distinct primes, $\displaystyle \phi(p^rq^s) = \phi(p^r)\phi(q^s)$
thus it suffices to find $\displaystyle \phi(p^r)$, where r is the highest power of p that divides n, for each prime p in the factorization of n.
by counting (never underestimate the power of counting), $\displaystyle \phi(p^r) = (p-1)(p^{r-1})$
so for n = 10, for example, 10 = (2)(5).
φ(2) = 1, φ(5) = 4, so φ(10) = φ(2)φ(5) = (1)(4) = 4.