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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?

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    $\begingroup$ Well, $103$ is a prime number so..... $\endgroup$ – fleablood Mar 16 at 15:05
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$\phi(n)$, 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

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

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As Euler stated

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

Thus, and since $103$ is prime

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

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