Perfect Square in 987654321...

First, let's consider what it means for $N^2$ to have starting digits of 987654321. This simply means that

$987654321 \times 10^n \leq N^2 < 987654322 \times 10^n$

While it's tempting to immediately take square roots, we don't really know how to deal with $10^ \frac{n}{2}$ when $n$ is odd. Instead, we consider the cases when $n$ is odd / even. This gives us:

$n \text{ odd: } 99380.7990006118 \ldots \times 10^k \leq N < 99380.7990509233 \ldots \times 10^k$ $n \text{ even: } 31426.9680529319 \ldots \times 10^k \leq N < 31426.968068418\ldots \times 10^k$

Now, we substitute increasing values of $k$ till we find the smallest integer $N$ that satisfies the inequality.

In the odd case:
When $k = 0,$ we get $99480.7990006118\ldots \leq N < 99380.7990509233 \ldots$ , which has no integer solution.
When $k = 1,$ we get $994807.990006118\ldots \leq N < 993807.990509233 \ldots$ , which has no integer solution.
When $k = 2,$ we get $9948079.90006118\ldots \leq N < 9938079.90509233 \ldots$ , which has no integer solution.
When $k = 3,$ we get $99480799.0006118\ldots \leq N < 99380799.0509233 \ldots$ , which has no integer solution.
When $k = 4,$ we get $994807990.006118\ldots \leq N < 993807990.509233 \ldots$ , which has no integer solution.
When $k =5,$ we get $9948079900.06118\ldots \leq N < 9938079905.09233 \ldots$ , which has several integer solutions, the smallest of which is 9938079901.

In the even case:
(do the same as above)
the smallest integer $N$ will be 3142696806. This is the answer.

There are only two options, since we can split $987654321\overline T$ in two different ways: 98'76'54'32'10'00'...'00 or 9'87'65'43'21'00'...'00.

(Fortunately, since there are three attempts to get the answer, but only two to choose from; we may do one and submit. If the former case fails, the latter will give the answer. If its do or die, do all cases first.)

In order to get the square root, consider $\overline{abc\ldots}^2 = (A+B+C+\ldots)^2= A^2 + 2AB + B^2 + 2AC + 2BC + C^2 + \ldots$ where $A = a \times 10^k,\; B = b \times 10^{k-1},\; C = c\times 10^{k-2}, \ldots$ for some $k \in \mathbb N$ .

Thus, $N^2 = \overline{abc\ldots}^2 = (A^2) + (2A+B)\cdot B + (2(A+B)+C)\cdot C + \ldots$ .

Observe that the split we made earlier corresponds to the RHS: the first term implies A = 9 (or 3); transpose the first term, then the second implies B = 9 (or 1); transpose, the second, then the third implies C = 3 (or 4); $\ldots$ . I'll leave the computations to you.

However the next problem would be when this seemingly infinite loop ends.

I'll demonstrate using with smaller numbers... say for 3 instead of 987654321. (I'll ignore case 30'00' $\ldots$ )

Here you can count the number of trailing zeros and the number of digits of the nth term factor. Observe that $17^2<N^2<18^2 = N^2 + B_F +B+1 - N_2 = 300 +27 + 7 + 1 - 11 = 324$ . The leading digit 3 is unaffected, therefore we can stop with a new $N^2 = 324$ and $N = 18$ .

Back to the question above, a similar scenario appears (latter case) when $N^2 > 9876543210000000000$ , and therefore, $N = 3142696806$ .

Without calculator:

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        9|87|65|43|21|  |  |  |  |  | ==> 3142696806
3 x 3 = 9            |
       ---           |
          87         |
 61 x 1 = 61         |
         ----        |
          26 65      |
624 x 4 = 24 96      |
         -------     |
           1 69 43   |
6282 x 2 = 1 25 64   |
          ---------  |
             43 79 21|
62846 x 6 =  37 70 76|
            ---------+
              6 08 45|00
 628529 x 9 = 5 65 67|61
             --------+---
                42 77|39 00
  6285386 x 6 = 37 71|23 16
               ------+------
                 5 06|15 84 00
  62853928 x 8 = 5 02|83 14 24
                -----+---------
                    3|32 69 76 00
    628539360 x 0 =  |          0
                  ---+------------
                    3|32 69 76 00 00
   6285393606 x 6 = 3|77 12 36 16 36
                   --+---------------
                    -|44 42 60 16 36 ...

In the same way I checked 98|76|54|32|1 | because of the slight possibility that it would reduce the "remainders" to less than 1 more quickly.

Since the given number has nine digits, we need at least ten zeros after it so that when the asked number is squared, all the digits of the given number retain their original positions. Thus extracting the square root of 9876543210000000000, we get 3142696805.29319, the integer next to which is 3142696806