Be a Wizard..Find the missing digits of this 41 digit Humongous Palindromic Square!

Question

Absolutely No calculators or computers.

Only Simplest Method for finding the missing digits will be accepted as the right answer.

Given:

The square of this 21 digit number

$ 229 359 782 235 085 482 225 $

is a 41 digit

Palindromic Square:

To emphasize the palindromic nature of the square, it is split as 2 lines. Top line shows first 20 digits and the swing digit 2 as the 21st. Second line gives the last 20 digits in reverse order.

$ 5xx xxx xxx 069 258 339 64 $ $2 $

$ 5xx xxx xxx 069 258 339 64 $

To avoid confusion, I am giving the full number from left to right also:

$5xx xxx xxx 069 258 339 64 $ $2$ $46 933 852 960 xxx xxx xx5$

Find the missing digits xxxxx...x can be any number from 0 to 9.

If you know the right method ...5 minutes or less.

Answer 1

The answer is (absolutely no counting devices, about 8 mins for counting + about 2x time for writing and formatting the answer :) )

26059097, i.e. $$229359782235085482225^2 = 52605909706925833964246933852960790950625$$

Explanation

The number ends with 25, so it should be of the form $200x + 25$, because there is an even digit before 25. We get $(200x+25)^2 = 40000x^2+10000x+625 = 10000(4x^2+x)+625$. It immediately means that the number ends with 0625 (so we get 260). To calculate the other 5 digits, note that $2x$ ends with 54822, so $x$ ends with 27411, and $4x^2=(2x)^2$ ends with 51684. The latter result as achieved by doing a partial multiplication keeping only the last 5 digits (it's easy to do):
54822
54822
--------
09644
9644
576
88
0
--------
51684
So, the other missing digits are 59097, reverse of $51684+27411=79095$.
P.S. We can find the (full) actual value of $x=1146798911175427411$ very quickly, but this is not needed for the solution.
P.P.S. The process can be optimised by using the $2000x+225$ representation in the 1st step to get 50625 at the end, and then multiplying 4-digit numbers.