If N² = 12345678987654321, then how many digits does the number N have ?
- (a) 8
- (b) 9 ✓ UPSC's answer
- (c) 10
- (d) 11
Why the answer is (b)
• The number 12345678987654321 has 17 digits.
• For a perfect square N², the number of digits in N is approximately half the number of digits in N², adjusted for the leading digit.
• Specifically, if a number has $2k$ or $2k-1$ digits, its square root has $k$ digits.
• Since 17 is an odd number of digits, we look at the range: $10^{16} \le 12345678987654321 < 10^{17}$.
• Taking the square root, $10^8 \le N < 10^{8.5}$.
• This implies N is at least 100,000,000 (9 digits) and less than 316,227,766 (9 digits).
• Therefore, N must have 9 digits.
Why the other options are wrong
- (a) 8
- An 8-digit number squared would result in a number with at most 16 digits, but the given number has 17 digits.
- (c) 10
- A 10-digit number squared would result in a number with at least 19 digits, which exceeds the 17 digits of the given number.
- (d) 11
- An 11-digit number squared would result in a number with at least 21 digits, which is far larger than the given 17-digit number.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2025, held on 25 May 2025. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.