2. Find the largest positive number which, after erasing one of the digits in its decimal representation, can double.
Solution. Answer: 0.375. It is obvious that when a digit in the integer part of a positive number is erased, the number decreases. Therefore, the digit that was erased must be from the fractional part. Let the first digit $x$ after the decimal point in the original number not be 0. Then, if a digit other than the first one after the decimal point is erased, the old number is no less than $n+0.1x$, the new one is less than $n+0.1(x+1)$, and the number increases by less than $\frac{n+0.1(x+1)}{n+0.1x}=1+\frac{0.1}{n+0.1x} \leqslant 1+\frac{1}{x} \leqslant 2$ times. From this estimate, it is clear that $n=0$ must hold. Thus, if the first digit after the decimal point is not 0, it must be erased (to satisfy the conditions). Let $y$ be the original number after replacing the first digit after the decimal point with 0. By the condition, $10y=2\left(\frac{x}{10}+y\right)$ or $x=40y$. Since $y<0.1$, we have $x<4$. For the maximum value of $x=3$, we get $y=\frac{3}{40}=0.075$. Therefore, the desired number is 0.375. This number is obviously greater than any number with the required property that has a 0 after the decimal point.