# How to calculate $\varphi(103)$?

How to calculate $$\varphi(103)$$? I know the answer is $$102$$ by looking at Wiki. But how can I find the multiplication of the prime numbers in order to use Euler's formula?

• Well, $103$ is a prime number so..... – fleablood Mar 16 at 15:05

$$\phi(n)$$, by definition, is the number of natural numbers less than $$n$$ that are relatively prime to $$n$$. And if $$p$$ is a prime number then all natural numbers less than $$p$$ are relatively prime to $$p$$, so $$\phi(p) = p-1$$.

$$103$$ is prime so $$\phi(103) = 103-1 = 102$$.

Furthermore if $$n = \prod p_i^{k_i}$$ is the unique prime factorization of $$n$$, then $$\phi(n) = \prod [(p_i-1)p_i^{k_i - 1}]=n\prod\limits_{p|n\\p\text{ is prime}}(1-\frac 1p)$$, which is how you would calculate any $$\phi(n)$$.

For example if $$\phi(104)$$, then as $$104 = 2^3*13$$, we'd have $$\phi(n) = [(2-1)2^{3-1}][(13-1)13^{1-1}]=[1\cdot2^2][12\cdot1]=4\cdot12=48$$. Or we could say $$\phi(n) = 104\cdot \prod\limits_{p|104}(1 - \frac 1p) = 104(1-\frac 12)(1-\frac 1{13})= (104 - 52)(1-\frac 1{13}) =52(1-\frac 1{13}) = 52 -4 =48$$.

For example

For each prime number $$p$$, $$\varphi(p)=p-1$$. And $$103$$ is a prime number.

As Euler stated

$$\varphi(n)=n·\prod_{p\mid n}\bigg(1-\frac{1}{p}\bigg)$$

Thus, and since $$103$$ is prime

$$\varphi(103)=103·\bigg(1-\frac{1}{103}\bigg)=103-1=102$$