$A$ says to $B$: "Take any number. Write its digits in reverse order and subtract the smaller number from the larger one. Multiply this difference by any number. In the resulting product, cross out any non-zero digit of your choice, and tell me the truncated number." $B$'s response: 35407, to which $A$ says that the crossed-out digit is 8.

How did $A$ know this?
The general form of a number in the decimal system is

$$
a_{0} 10^{n}+a_{1} 10^{n-1}+\ldots+a_{n-2} 10^{2}+a_{n-1} 10+a_{n}
$$

which can also be written as

$$
\begin{gathered}
a_{0}\left(10^{n}-1\right)+a_{1}\left(10^{n-1}-1\right)+\ldots+a_{n-2}\left(10^{2}-1\right)+a_{n-1}(10-1)+ \\
+a_{0}+a_{1}+a_{2}+\ldots+a_{n-1}+a_{n}
\end{gathered}
$$

Since $10^{k}-1$ is divisible by $10-1=9$ for any non-negative integer value of $k$, when we divide a number by 9, we get the same remainder as when we divide the sum of its digits by 9. Therefore, if a number is of the form $9 n+k(k=0,1,2, \ldots, 8)$, then the number obtained by reversing the order of its digits is of the form $9 n^{\prime}+k$, and thus the difference between the two numbers: $9\left(n-n^{\prime}\right)$ is always divisible by 9. If a number divisible by 9 is multiplied by any number, the product is also divisible by 9.

$A$ can determine the crossed-out digit by finding the sum of the digits in the number $B$ states $(3+5+4+0+7=19)$, and then rounding this sum to the nearest number divisible by 9 $(19+8=27)$. If the number $B$ states is divisible by 9, then the crossed-out digit is 9, because it was stipulated that 0 cannot be crossed out.

Note: It is evident from the above solution that it is sufficient for the removed digit to be composed of the same digits as the original number for the difference to be divisible by 9.

Lassányi Ferenc (Bp, VIII. Piarist g. II. o. t.)