If N² = 12345678987654321, then how many digits does the number N have?

Correct Option

The given number N² = 12345678987654321 exhibits a specific pattern: it sequentially increases from 1 to 9 and then decreases back to 1. This form is characteristic of the square of a repunit, which is a number consisting solely of the digit 1.

Specifically, if a repunit consists of 'k' digits (i.e., 'k' ones), its square will form a number that ascends from 1 to 'k' and then descends back to 1. For example:

• 1² = 1 (k=1)
• 11² = 121 (k=2)
• 111² = 12321 (k=3)

In the given N² = 12345678987654321, the highest digit reached in the sequence is 9. This indicates that N is a repunits composed of nine 1s.

Therefore, N = 111,111,111, which is a number with 9 digits.

Incorrect Options

Options 1 (8), 3 (10), and 4 (11) are incorrect because they do not align with the observed pattern of N².

• If N had 8 digits (i.e., N = 11,111,111), then N² would be 123456787654321, where the peak digit is 8. This does not match the given N².
• If N had 10 digits (i.e., N = 1,111,111,111), then N² would be a number with 19 digits, where the sequence would theoretically ascend to 10. The given N² has 17 digits and peaks at 9.
• If N had 11 digits (i.e., N = 11,111,111,111), then N² would be a number with 21 digits, which is inconsistent with the given N².

The unique structure of N² = 12345678987654321 directly corresponds to the square of a 9-digit repunit.