The reason is that square root depends critically on such pairs . You can find the some of the details at [this answer.](https://socratic.org/questions/while-finding-root-of-a-square-number-in-dividing-method-why-we-make-double-of-t#389218)
For example, let us consider squares of three digit numbers starting from 1,2,3,4,5,6,7,8,1,2,3,4,5,6,7,8, and 99.
These will be as follows (here pqpq represent digits in tens and units place):
(1pq)^2(1pq)2 ranges from 1,00,001,00,00 to 3,99,993,99,99
(2pq)^2(2pq)2 ranges from 4,00,004,00,00 to 8,99,998,99,99
(3pq)^2(3pq)2 ranges from 9,00,009,00,00 to 15,99,9915,99,99
(4pq)^2(4pq)2 ranges from 16,00,0016,00,00 to 24,99,9924,99,99
(5pq)^2(5pq)2 ranges from 25,00,0025,00,00 to 35,99,9935,99,99
(6pq)^2(6pq)2 ranges from 36,00,0036,00,00 to 48,99,9948,99,99
(7pq)^2(7pq)2 ranges from 49,00,0049,00,00 to 63,99,9963,99,99
(8pq)^2(8pq)2 ranges from 64,00,0064,00,00 to 80,99,9980,99,99
(9pq)^2(9pq)2 ranges from 81,00,0081,00,00 to 99,99,9999,99,99
Now assume a number like 8352183521, which is 289^22892
If we make pairs from left and select 8383, we will considering a number starting from 99 and searching for p's and q's and this will make us very much away from the real square root.
But not so, if we make pairs from right and think of a number like 2pq2pq.
