Probability.

If a $10$ -digit number (with digits $0,1,2,3,4,5,6,7,8,9$ ) is to be divisible by $11$ , then the digits $0,1,2,3,4,5,6,7,8,9$ must be divided into two sets of five numbers (the digits occuring in positions 1,3,5,7,9 in one set, and the digits occuring in positions 2,4,6,8,10 in the other). If $A,B$ are the digit sums of the two sets, then $A+B = 45$ and $A-B$ is a multiple of $11$ . Since $A,B$ are integers, $A-B$ is odd. If $|A-B| \ge 55$ , then one of $A,B$ is negative. If $|A-B| = 33$ , then one of $A,B$ is equal to $6$ , which is impossible. Thus we must have $|A-B|=11$ , and hence $\{A,B\} = \{17,28\}$ .

There are precisely $11$ subsets of $\{0,1,2,3,4,5,6,7,8,9\}$ whose total is $11$ . For each such set $S$ , let $T = \{0,1,2,3,4,5,6,7,8.9\}\backslash S$ . For each such set $S$ , then either the digits of $S$ form the digits in positions $1,3,5,7,9$ , or else the digits of $T$ do. There are $5!$ ways of arranging any set of $5$ numbers, and there are $5! - 4! = 4 \times 4!$ ways of arranging $5$ numbers so that one particular number ( $0$ ) does not appear first in the list. Thus there are $5! \times 5! + 4 \times 4! \times 5! \; = \; 9 \times 4! \times 5!$ ways of arranging the digits in a particular pair of sets $S$ and $T$ to obtain a $10$ -digit number. Thus there are $11 \times 9 \times 4! \times 5!$ $10$ -digit numbers comprising the digits $0,1,2,3,4,5,6,7,8,9$ that are divisible by $11$ .

There are $10! - 9! = 9 \times 9!$ $10$ -digit numbers comprising the digits $0,1,2,3,4,5,6,7,8,9$ . Thus the desired probability is $\frac{11 \times 9 \times 4! \times 5!}{9 \times 9!} \; =\; \frac{11 \times 4!}{9 \times 8 \times 7 \times 6} \; = \; \frac{11}{9 \times 7 \times 2} \; = \; \frac{11}{126}$ making the answer $11 + 126 = \boxed{137}$ .

Let the number $n$ be represented $\overline{abcdefghil}$ where each letter represents a digit. Then $n$ is divisible by $11$ if and only if the “odd-digit” sum $a+c+e+g+i$ and the “even-digit” sum $b + d + f + h + l$ differ by a multiple of $11$ . However, since we know that:

$a + b + c + d + e + f + g + h + i + l= 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45,$

the only possibility is that one sum (the odds or the evens) is $17$ and the other is $28$ . We can now list the sets of 5 distinct digits that sum to 28 (the other 5 digits will necessarily have to sum to 17):

$9+8+7+4+0$

$9+8+7+3+1$

$9+8+6+5+0$

$9+8+6+4+1$

$9+8+6+3+2$

$9+8+5+4+2$

$9+7+6+5+1$

$9+7+6+4+2$

$9+7+5+4+3$

$8+7+6+5+2$

$8+7+6+4+3$

Thus, there are 11 ways to split the digits into one group of five that sum to 28 and one group of five that sum to 17. For each of these 11 ways, we have 2 choices which digits get put into the odd positions and which get put into the even positions, and then 5! ways to arrange the five digits in each group. However, exactly $\frac{1}{10}$ of these will end up with 0 as the leading digit, which is not allowed.

Therefore, there are $11\times 2\times (5!)^2 \times\frac{9 }{10} =\frac{ 99 }{5}\times(5!)^2$ numbers in S that are divisible by $11$ . There are $(9 · 9!$ ) numbers in $S$ , so the probability is:

$\frac{(99)(5!)^2}{5\times9\times 9!}=\frac{11\times4!}{9\times 8\times7\times6}=\frac{11}{126}\implies 11+126=137$