Euler Phi Function of 25,935
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Example of Euler $\phi$ Function of Square-Free Integer
- $\map \phi {25 \, 935} = 10 \, 368$
where $\phi$ denotes the Euler $\phi$ Function.
Proof
From Euler Phi Function of Square-Free Integer:
- $\ds \map \phi n = \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$
where $p \divides n$ denotes the primes which divide $n$.
We have that:
- $25 \, 935 = 3 \times 5 \times 7 \times 13 \times 19$
and so is square-free.
Thus:
\(\ds \map \phi {25 \, 935}\) | \(=\) | \(\ds \paren {3 - 1} \paren {5 - 1} \paren {7 - 1} \paren {13 - 1} \paren {19 - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \times 4 \times 6 \times 12 \times 18\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \times 2^2 \times \paren {2 \times 3} \times \paren {2^2 \times 3} \times \paren {2 \times 3^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2^7 \times 3^4\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 10 \, 368\) |
$\blacksquare$