Think outside the box

Let the integer formed by this digits is $N=\overline{abcdefg}$

To be divisible by $11$ the below condition should satisfy

$(a+c+e+g)-(b+d+f)=11k$

Maximum $k$ can occur $k=0,1$ because $(7+6+5+4)-(1+2+3)=16 <22$ .So, $0,11$ are only possible values

Case 1 $k=0$

$a+c+e+g=b+d+f \\ a+b+c+d+e+f+g=2(b+d+f)=28 \\ b+d+f=14$ .

Let $b=6$ then $d+f=8$ .Possible pairs are $(1,7);(3,5)$

Let $b=7$ then $d+f=7$ .Possible pairs are $(1,6);(2,5);(3,4)$

So, distinct triplets are $(1,6,7);(3,5,6);(2,5,7);(3,4,7)$

If $b+d+f$ is evaluated then automatically $a+c+e+g$ will be same.

So, for each triplet number of positive integers are $4! \times 3! \times 4=576$

Case 2 : $k=1$

Here $a+c+e+g=11+b+d+f \\ a+b+c+d+e+f+g=11+2(b+d+f) \\ 2(b+d+f)=17$ .

So, no integral solution exists.

Therefore the probability is $\dfrac{576}{7!}=\dfrac{4}{35}$

The alternating sum of the digits of $63195$ 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, $63195$ 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 $9$ .

Let the alternating sum of the digits be $S$ , and the 7 digit numbers' digits be $a,b,c,d,e,f,g$ . Then, $S=a-b+c-d+e-f+g$ . We can group the alternating sum differently in a way that the digits that contribute positively to $S$ and the digits that contribute negatively to $S$ are separated: $S=(a+c+e+g)-(b+d+f)$ . A 7 digit integer always contains 4 digits that contribute positively to $S$ and 3 that contribute negatively.

Let the sum of the digits that contribute positively to $S$ be $P$ , and the digits that contribute negatively be $N$ ; $P=a+c+e+g$ and $N=b+d+f$ . Ergo, it follows that $S=(a+c+e+g)-(b+d+f)=P-N$ .

We determine the largest possible alternating value of $S$ by choosing the 4 largest integers to make up $P$ (that is, 4, 5, 6, 7), and the three smallest integers (1 ,2, 3) to make up $N$ ; the largest possible alternating sum is: $S=(4+5+6+7)-(1+2+3)=16$ .

On the other hand, we choose the smallest possible alternating value of $S$ by selecting the 4 smallest integers (1, 2, 3,4) to equal $P$ and the 3 largest integers (5, 6, 7) to make up $N$ . Thus the smallest possible value of $S=(1+2+3+4)-(5+6+7)=-8$ . Since $S$ must be divisible by $11$ , and is between $-8$ and $16$ , then either $S=11$ or $S=0$ . We should also note that the sum of the first 7 positive integers is $1+2+3+4+5+6+7=28$ , and since each of these 7 integers must contribute to either $P$ or to $N$ , then $P+N=28$ .

Case 1: The alternating sum of the digits is 11, or $S=11$

If $S=11$ , then $S=P-N=11$ . Since 11 is an odd number, then either $P$ is even and $N$ is odd, or the other way around. In other terms, $P$ and $N$ must have different parity. However, if either one of $P$ or $N$ is even and the other is odd, then their sum must be an odd number. But we know that $P+N=28$ , an even number. Therefore, it is not possible that $S=11$ .

Case 2: The alternating sum of the digits is 0, or $S=0$

If $S=0$ , then $P-N=0$ , and so $P=N$ . Since $P+N=28$ , then $P=N=14$ .

We find all groups of 3 digits, chosen from the digits 1 to 7, such that their sum $N=14$ .There are exactly 4 possibilities: (7, 6,1), (7, 5, 2), (7, 4, 3), and (6, 5, 3). In each of these 4 cases, the digits from 1 to 7 that were not chosen, (2, 3, 4, 5), (1, 3, 4, 6), (1, 2, 5, 6), and (1, 2, 4, 7), respectively, represent the 4 digits whose sum is $P=14$ .

There is a summarization of this in the table below:

Consider the first row of numbers in this table above. Each arrangement of the 4 digits 2, 3, 4, 5 combined with each arrangement of the 3 digits 7, 6, 1 (in the required way) gives a new 7-digit integer whose alternating digit sum is $0$ . Two such arrangements are shown (you may check that $S = 0$ for each). Since there are $4 × 3 × 2 × 1 = 24$ ways to arrange the 4 digits ( $4$ choices for the first digit, $3$ choices for the second, $2$ choices for the third and $1$ choice for the last digit), and $3 × 2 × 1 = 6$ ways to arrange the 3 digits, then there are $24 × 6 = 144$ ways to arrange the 4 digits and the 3 digits. Each of these 144 arrangements is different from the others, and since $P = N = 14$ for each, then $S = P - N = 0$ and so each of the 144 7-digit numbers is divisible by 11. Similarly, there are also 144 arrangements that can be formed with each of the other 3 groups of integers that are shown in the final 3 rows of the table. That is, there are a total of $144 × 4 = 576$ 7-digit integers (formed from the integers 1 through 7) which are divisible by 11. The total number of 7-digit integers that can be formed from the integers 1 through 7 is equal to the total number of arrangements of the integers 1 through 7, or $7×6×5×4×3×2×1 = 5040$ .

Therefore, when the digits 1 through 7 are each used to form a random 7-digit integer, the probability the number formed is divisible by 11 is $\dfrac{576}{5040}=0.114$ .

Let N be our integer with A being the sum of the digits in the odd positions (1st,3rd,5th, 7th positions) and let B be the sum of the digits in the even positions. Then we require 11| A-B , that is, $\ A-B = 11k \quad for \ k \in \mathbb{Z}$

Now 1+2+3 = 6 and 7+6+5+4 = 2 so $\ 6 \leq A,B \leq 22 \quad \therefore |A-B| \leq 16$ . Hence: $\ A-B = 0, \pm 1$ . Lets consider these two cases.

Case1: $\ k = \pm 1$

We now have two equations in terms of A and B, (which we can easily solve) namely $\ A+B = 28 \quad A-B = \pm 11 \implies 2A = 39 \ or \ 2B = 39$ So this case gives no solutions as it is impossible.

Case2: k= 0

Here we have A=B = 14 giving 4 triples for B namely, (7,6,1),(7,5,2),(7,4,3) and (6,5,3). For each triple, there are 3! arrangements and 4! arrangements for the corresponding quadruple A giving $\boldsymbol{3!4!}$ arrangements for each of $\boldsymbol{4}$ triples. There are 7! arrangements of 1,2,...,7 so (denoting our integer as N) we have: $\mathbb{P} (11|N) = \frac{4 \times 4!3!}{7!} = \frac{4}{35}$