Permutation of distinct digits

The rule for divisibility by $11$ is that the alternating sum of an integer be divisible by $11$ . As an example, the alternating sum of $8239$ is $8 - 2 + 3 - 9 = 0$ , which is divisible by $11$ and thus so is $8239$ , (indeed $8239 = 11 \times 749$ ). Now the sum of the digits $1$ through $8$ is $36$ , so one way we can create a number that is divisible by $11$ is to choose 4 digits that sum to $18$ , leaving the remaining 4 digits to sum to $18$ as well, and then alternating these two sets of digits. So choosing $1,4,5,8$ as one 4-digit set, we know by the alternating-sum rule that $12435687$ will be divisible by $11$ , since $1 - 2 + 4 - 3 + 5 - 6 + 8 - 7 = 0$ . The answer is therefore $\boxed{\text{Yes}}$ .

Note: $12435687 = 11 \times 1130517$ . A potential follow-up question would be to find the probability that a random permutation of $12345678$ is divisible by $11$ .