3. If $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{s}^{\alpha_{1}}$, then
$$
\varphi(n)=n\left(1-\frac{1}{p_{1}}\right)\left(1-\frac{1}{p_{2}}\right) \cdots\left(1-\frac{1}{p_{s}}\right) \text {. }
$$

Find the smallest positive integer $n$, such that $\varphi(n)=\frac{2^{5}}{47} n$.
3. From the problem, we have
$$
47\left(p_{1}-1\right)\left(p_{2}-1\right) \cdots\left(p_{s}-1\right)=2^{5} p_{1} p_{2} \cdots p_{s} \text {. }
$$

Thus, one of \( p_{1}, p_{2}, \cdots, p_{s} \) must be 47. Without loss of generality, let \( p_{1}=47 \). Then
$$
\begin{array}{l}
46\left(p_{2}-1\right)\left(p_{3}-1\right) \cdots\left(p_{s}-1\right)=2^{5} p_{2} p_{3} \cdots p_{s} \\
\Rightarrow 23\left(p_{2}-1\right)\left(p_{3}-1\right) \cdots\left(p_{s}-1\right) \\
\quad=2^{4} p_{2} p_{3} \cdots p_{s} .
\end{array}
$$

Without loss of generality, let \( p_{2}=23 \). Then
$$
\begin{array}{l}
22\left(p_{3}-1\right)\left(p_{4}-1\right) \cdots\left(p_{s}-1\right)=2^{4} p_{3} p_{4} \cdots p_{s} \\
\Rightarrow 11\left(p_{3}-1\right)\left(p_{4}-1\right) \cdots\left(p_{s}-1\right) \\
\quad=2^{3} p_{3} p_{4} \cdots p_{s} .
\end{array}
$$

Without loss of generality, let \( p_{3}=11 \). Then
$$
5\left(p_{4}-1\right)\left(p_{5}-1\right) \cdots\left(p_{s}-1\right)=2^{2} p_{4} p_{5} \cdots p_{s} \text {. }
$$

Without loss of generality, let \( p_{4}=5 \). Then
$$
\left(p_{5}-1\right)\left(p_{6}-1\right) \cdots\left(p_{s}-1\right)=p_{5} p_{6} \cdots p_{s} \text {. }
$$

Therefore, the smallest \( n = 47 \times 23 \times 11 \times 5 \).