Think Smart

I solved it by the simple and obvious, Principle of counting ;) :D

Let us first find all the 7 digit numbers, the sum of whose digits is 59.

After doing some reasoning, you'll be able to figure out that the least number of * '9's * possible for such a combination is 3 .

As 59=9+9+9+8+8+8+8 . ( Try removing one 9 and replacing it with any other digit. It's not possible. )

For this particular combination- total permutations= 7! / (3!*4!) = 35

similarly, 59=9+9+9+9+8+8+7 . total permutations= 7! / (2!*4!) = 105

59=9+9+9+9+9+8+6 . --> 7! / 5! = 42

59=9+9+9+9+9+7+7 --> 7! / (5!*2!) =21

and lastly, 59=9+9+9+9+9+9+5 --> 7

No other combination of digits is possible. So, the total number of numbers whose sum of digits is 59 is 35+105+42+21+7 = 210

Now we will find those numbers out of these which are divisible by 11.

consider a 7 digit number _ _ _ _ _ _ _ . It will be divisible by 11 only if the difference between the sum of odd place digits and sum of even place digits is divisible by 11 . (According to the divisibility rule taught at elementary level)

Here, there will be 4 odd place digits i.e. 1st, 3rd, 5th and 7th digit whereas 3 even place digits.

Let the sum of odd place digits be A and that of even place digits be B . Now, A+B=59 (obvious)

and A-B (or B-A) =0, 11, 22, 33 etc. (Since difference has to be divisible by 11)

Also, values greater than 22 can be discarded for A-B ( Think it over. Can it ever be possible? ).

Now, we are left with A-B=0 or 11 or 22

But if A-B=0 or 22 , THEN using A+B=59 results in non integral values of A and B which is clearly not possible.

Thus A-B ( or B-A , as we do not as of yet, know which is greater ) can only be equal to 11 .

this gives, A=35 and B=24

But since A is the sum of 3 digits it cannot be greater than 9+9+9=27 (think it over). Thus , A=24 and B=35

Now, let us reason out the number of combinations possible for A=24

It is not very tough to see that -->

CASE 1: 24=8+8+8.

For this combination only one combination out of the total 5 ( given at the top ) for the sum to be 59 , is possible.

That is, 59=9+9+9+8+8+8+8

Hence, the number must be of the form _ 8 _ 8 _ 8 _ , where the remaining digits are to be filled with 9,9,9 & 8

Since, the 4th 8 can fill any of the FOUR blanks, total number of permutations= 4

CASE 2: 24=9+9+6

For this combination, the possible combination of 7 digits is, 59=9+9+9+9+9+8+6 ( Referring to the pre-determined 5 combinations )

The number must be of the form _ 9 _ 9 _ 6 _

OR _ 9 _ 6 _ 9 _

OR _ 6 _ 9 _ 9 _

Thus, the digit 6 can fill any one of the 3 blanks ,while the digit 8 can fill any one of the remaining 4 blanks . Each instance will be a different permutation . (We have not included the '9's in our calculation as they are all same and will fill the remaining 5 places, not affecting the no. of permutations.)

So, total permutations for this case= 4 * 3 = 12

CASE 3: 24= 9+8+7

The possible combination of 7 digits for this one is 59 = 9+9+9+8+8+7

Again, As we have proceeded for the above TWO , the number must be something like _ _ _ _ _ _ _ with 8 & 7 occupying two of the 3 odd places , the rest FOUR occupied by 9,9,9,8

Thus, 8 & 7 can occupy any two of the three odd places while the remaining digit 8 can fill any one of the 4 places.

So, total number of permutations= (3!*2!) * 4 = 24

Since, NO other case is possible , we add the total number of permutations which are divisible by 11 . --> 4+12+24 = 40

Finally, the probability that the 7 digit number whose sum of digits is divisible by 11 will be

(Permutations divisible by 11) / (Total number of permutations) $= \frac {40}{210} = \frac {4}{21}$ Hence, answer $= 4 + 21 = \boxed{25}$

Let it be $n=d_1 d_2...d_7$ the digits representation of the number. We know that

$\displaystyle \sum_{i=1}^{7} d_i=59, \ d_1 \geq 1, \ d_i \leq 9, i \in [2,9]$ .

In order to be divisible by $11$ , we need that

$\displaystyle d_1+d_3+d_5+d_7-(d_2+d_4+d_6) \equiv 0 \mod{11}$ .

Let's call $a=d_1+d_3+d_5+d_7$ and $b=d_2+d_4+d_6$ . So, we have that

$\begin{cases} a+b=59 \\ a-b=11k, \ k \in \mathbb{N} \end{cases}$

$\begin{cases} \displaystyle a=\frac{59+11k}{2} \\[7pt] \displaystyle b=\frac{59-11k}{2} \end{cases}$

Since $1 \leq a \leq 36$ and $0 \leq b \leq 27$ , than

$\begin{cases} -5 \leq k \leq 1 \\ 1 \leq k \leq 5 \end{cases} \ \Longrightarrow \ k=1 \ \Longrightarrow \ \begin{cases} a=35 \\ b=24 \end{cases}$ .

The number of $7$ -digits numbers such that their sum is $59$ is equal to the number of solutions $|S_T|$ of

$\displaystyle S_T=\sum_{i=1}^{7} d_i=59, \ d_i \leq 9, \ i \in [1,7]$ ,

while the ones that also satisfy $n \equiv 0 \mod{11}$ are $|S_1||S_2|$ , where $|S_k|$ , $k=[1,2]$ are the number of solution of

$\displaystyle S_1=\sum_{i=0}^{3} d_{2i+1} = 35 \ \text{and} \ S_2=\sum_{i=1}^{3} d_{2i} = 24$ .

In order to find $|S_h|$ , $h=\{T,1,2\}$ we use a classical star and bars argument:

$\displaystyle S_T=\sum_{i=1}^{7} d_i=59, \ d_i \leq 9 \\ y_i=9-d_i \geq 0 \\ \sum_{i=1}^{7} y_i=9 \cdot 7 - 59=4$

hence,

$\displaystyle |S_T|=\binom{4+7-1}{7-1}=\binom{10}{7}=210$

With the same process we get $\displaystyle |S_1|=\binom{1+4-1}{4-1}=\binom{4}{3}=4$ and $\displaystyle |S_2|=\binom{3+3-1}{3-1}=\binom{5}{2}=10$ . Eventually, our probability is

$\displaystyle \mathbb{P}\bigg(n \equiv 0 \mod{11} \bigg| \sum_{i=1}^{7} d_i=59\bigg)=\frac{|S_1||S_2|}{|S_T|}=\frac{4}{21} \\ 4+21=\boxed{25}$