Combinatorics is making it...

SOLUTION

Check the divisibility of 9 first, You will get :-

Lets see what could be the largest and smallest sum of digits we can get from a seven digit number.

Largest number : $9876543,$ giving sum of $9 + 8 + 7 + 6 + 5 + 4 + 3 = 42$ .

Smallest number : $1234567,$ giving a sum of $28$ .

Thus we can only get $36$ as a multiple of $9$ without any other possibility,

so the combinations would be 1236789, 12 45 789 , 1 3 45 6 89, 2 3456 7 9. BOLD is showing how I altered the digits.

Now, Think of Seven places as positions where you need to place digits _ _ _ _ _ _ _ .

Last digit could not hold $2,4,6,8$ and if you see carefully then there are exactly $3$ even digits in all the numbers. Hence it's permutations that would not be divisible by $2$ will be

$6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 4 = 6! \times 4$ . Hence at last postion only 4 digits could come.

Thus in all $4 \times 6! \times 4 = 11520$ .

But the story doesn't end here, $11520 + 12^2 = 11664 \ and \ \sqrt {11664} = 108 .$ This is the answer !