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It might seem counterintuitive, but if you convolute these complicated things, they'll simplify in no time! Extend your understanding of the totient function by studying this class of functions.

If \( f: \mathbb{N}\mapsto \mathbb{N} \) is a bijective function that satisfies

\[ f(xy ) = f(x) f(y) \]

and \( f(2015) = 42 \), what is the minimum value of \( f(2000) \)?

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\[\sum_{d|n}\mu\left(\frac{n}{d}\right)f(d)=n\]

If \(f(d)\) is an arithmetic function such that the equation above holds for all positive integers \(n\), find \(f(2015)\).

**Notation**

- \(\mu\) denotes the Möbius function.

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\[\sum_{\mu(n)=1} \frac{1}{n^2} = \frac{A}{B\pi^C}\]

Let \(\mu(n)\) denote the möbius function, the sum is taken over all positive integers \(n\) such that \(\mu(n)=1\), with coprime positive integers \(A\) and \(B.\) Find \(A+B+C\).

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Compute \[ \large \sum_{d|2015!}\mu(d)\phi(d).\]

**Notations**

\(\mu(d)\) denotes the Möbius function.

\(\phi(d)\) denotes Euler's totient function.

\(d\) are the positive factors of \(2015!.\)

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