116. Once again about division by 37. Here is an interesting development of the previous puzzle. Nine digits $1,2,3,4,5,6,7,8,9$ are written in a random order, for example 412539768. What is the probability that the resulting number is divisible by 37 without a remainder?
116. Let's denote our number as \(A B C A B C A B C\). If the sums of the digits denoted by the letters \(A, B\), and \(C\) are respectively:

\begin{tabular}{rrr} 
A & B & C \\
18 & 19 & 8 \\
15 & 15 & 15 \\
12 & 11 & 22 \\
19 & 8 & 18 \\
22 & 12 & 11 \\
8 & 18 & 19 \\
11 & 22 & 12
\end{tabular}

then in the first three cases \(11 A - 10 B = C\), in the next two \(11 A - 10 B - C = 111 (3 \times 37)\). Finally, in the last two cases \(10 B + C - 11 A = 111\). If any of these cases holds, regardless of the specific values of the corresponding digits, our number is divisible by 37. Here is an example of the first case:

\[
\begin{array}{ccccccccc}
A & B & C & A & B & C & A & B & C \\
9 & 8 & 4 & 7 & 6 & 3 & 2 & 5 & 1
\end{array}
\]

where the sum of the \(A\)-digits is 18, the \(B\)-digits is 19, and the \(C\)-digits is 8.

It is easy to see that the first three cases can occur in 22 ways, the next two in 10 ways, and the last two in 10 ways, making a total of 42 cases. In each case, the number of permutations of the \(A\) digits is 6, the \(B\) digits is 6, and the \(C\) digits is also 6. The total number of permutations is \(6 \times 6 \times 6 = 216\). Multiplying the number of cases by the number of permutations, we get 9072 favorable (the number is divisible by 37) outcomes. Since the number of permutations of nine digits is 362880, the probability of a favorable outcome is \(\frac{9072}{362880}\), or \(\frac{1}{40}\). In other words, there are 39 chances against 1 that the number will not be divisible by 37.