Euler Phi Function of 25,935

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:

\(\ds \map \phi {25 \, 935}\)

\(=\)

\(\ds \paren {3 - 1} \paren {5 - 1} \paren {7 - 1} \paren {13 - 1} \paren {19 - 1}\)

\(\ds \)

\(=\)

\(\ds 2 \times 4 \times 6 \times 12 \times 18\)

\(\ds \)

\(=\)

\(\ds 2 \times 2^2 \times \paren {2 \times 3} \times \paren {2^2 \times 3} \times \paren {2 \times 3^2}\)

\(\ds \)

\(=\)

\(\ds 2^7 \times 3^4\)

\(\ds \)

\(=\)

\(\ds 10 \, 368\)

$\blacksquare$