56. Probability of divisibility. Find the probability that if the digits $0,1,2, \ldots, 9$ are placed in random order in the empty spaces in the sequence of digits

$$
5-383-8-2-936-5-8-203-9-3-76
$$

then the resulting number will be divisible by 396.
56. The number 76, formed by the last two digits, is divisible by 4. The difference between 73 (the sum of all digits in even positions) and $17+45$ (the sum of all digits in odd positions) is divisible by 11 regardless of the order in which the empty places are filled*. The sum of all digits, $90+45$, is divisible by 9. Therefore, our number is divisible by $4 \cdot 11 \cdot 9=396$. Hence, the desired probability is 1.

[P. Nagara, A. M. M., 58, 700 (December 1951).]