If N^2 = 12345678987654321, then how many digits does the number N have?

Question

Accepted Answer

The correct answer is option B. This question relies on recognizing a specific pattern of squares of numbers consisting only of the digit '1'.
Let's observe the pattern:
- 1^2 = 1 (N has 1 digit, N^2 has 1 digit, peak digit in N^2 is 1)
- 11^2 = 121 (N has 2 digits, N^2 has 3 digits, peak digit in N^2 is 2)
- 111^2 = 12321 (N has 3 digits, N^2 has 5 digits, peak digit in N^2 is 3)
- 1111^2 = 1234321 (N has 4 digits, N^2 has 7 digits, peak digit in N^2 is 4)

The pattern shows that if N consists of 'k' number of '1's, then N^2 will be a number that increases from 1 to 'k' and then decreases back to 1. The peak digit in N^2 directly corresponds to the number of '1's in N.

In the given N^2 = 12345678987654321, the highest digit (the peak) is 9.
Therefore, N must be a number consisting of nine '1's (i.e., N = 111,111,111).
Thus, the number N has 9 digits. This question assesses the ability to identify numerical patterns and apply them to solve problems, a common feature in UPSC CSAT's quantitative aptitude section.

Q10. If N^2 = 12345678987654321, then how many digits does the number N have?

Detailed Solution

More Basic Numeracy Questions

More from UPSC 2025 CSAT

Ace UPSC with AI-powered Practice