2015 Gauss 8 Problem 25

The *alternating sum* of the digits of 63\,195 is 6 - 3 + 1 - 9 + 5 = 0. In general, the alternating sum of the digits of a positive integer is found by taking its leftmost digit, subtracting the next digit to the right, adding the next digit to the right, then subtracting, and so on. A positive integer is divisible by 11 exactly when the alternating sum of its digits is divisible by 11. For example, 63\,195 is divisible by 11 since the alternating sum of its digits is equal to 0, and 0 is divisible by 11. Similarly, 92\,807 is divisible by 11 since the alternating sum of its digits is 22, but 60\,432 is not divisible by 11 since the alternating sum of its digits is 9.

Lynne forms a 7-digit integer by arranging the digits 1, 2, 3, 4, 5, 6, 7 in random order. What is the probability that the integer is divisible by 11?

\textbf{(A)}\ \frac{1}{35}\quad \textbf{(B)}\ \frac{5}{42}\quad \textbf{(C)}\ \frac{3}{35}\quad \textbf{(D)}\ \frac{1}{42}\quad \textbf{(E)}\ \frac{4}{35}

If there are no answer choices shown, enter a numerical answer.

Full credit to this problem is given to the CEMC, you may view all Gauss contests here.