Digits in N where N-squared is a palindrome

The given square number 12345678987654321 exhibits a perfect ascending and descending sequence peaking at 9. This is a well-known mathematical pattern for squares of numbers consisting entirely of ones (repunits). 11² = 121 (max digit 2, length 2) 111² = 12321 (max digit 3, length 3) Following this pattern, a number peaking at digit k is the square of a number made of exactly k ones. Here, the peak digit is 9, so N must be 111111111, which is exactly 9 digits long.

Answer: (b).