3-2. In the decimal representation of a positive number \(\alpha\), all decimal places starting from the third digit after the decimal point are discarded (i.e., an approximation of \(\alpha\) with a deficiency is taken to an accuracy of 0.01). The resulting number is divided by \(\alpha\), and the quotient is again rounded down to the same accuracy. What numbers can result from this (list all possible values)?
Solve 2. Answer: \(0, \frac{50}{100}, \frac{51}{100}, \ldots, \frac{99}{100}, 1\).

Let \(\alpha=\frac{n}{100}+\alpha_{1}\), where \(n\) is an integer and \(0 \leqslant \alpha_{1}0\), then \(\frac{n}{n+100 \alpha_{1}}>\frac{1}{2}\), since \(100 \alpha_{1}<1\). The fraction \(\frac{n}{n+100 \alpha_{1}}\) can take all values from \(1 / 2\) to 1. Indeed, let \(n=1\). As \(\alpha_{1}\) changes from 0 to \(\frac{1}{100}\), the number \(\frac{1}{1+100 \alpha_{1}}\) changes from 1 to \(1 / 2\).