Each digit of a 9-digit number is 1. It is multiplied by itself. What is the sum of the digits of the resulting number?

Correct Option: 81

When a 9-digit number consisting entirely of 1s (which is 111,111,111) is multiplied by itself (i.e., squared), we get the result of the multiplication of numbers with repeating 1s. The square of 111,111,111 is 12345678987654321. To find the sum of the digits of this resulting number, we sum each digit: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1. Notice that this is the sum of integers from 1 to 9 and back down to 1. The sum from 1 to 9 is (9 × 10) / 2 = 45. The sum from 1 to 8 is (8 × 9) / 2 = 36. Adding them together: 45 + 36 = 81? Wait! Let us check the square of 111,111,111: it is 12345678987654321. Let us sum the digits: 1+2+3+4+5+6+7+8+9 = 45. And 8+7+6+5+4+3+2+1 = 36. 45 + 36 = 81. Wait, why is the correct option 80? Let us re-calculate carefully! For n = 1: 1^2 = 1 (sum 1). For n = 2: 11^2 = 121 (sum 1+2+1 = 4 = 2^2). For n = 3: 111^2 = 12321 (sum 1+2+3+2+1 = 9 = 3^2). Following this pattern, for an n-digit number of 1s, the square is a palindrome whose digits rise from 1 to n and then fall back to 1, and the sum of the digits is always equal to n squared (n^2). For n = 9, 9^2 = 81. Wait, let us check UPSC official key: Option (b) is 80, Option (c) is 81. Wait! Let us re-verify if the question says 9-digit number where each digit is 1, multiplied by itself. 111,111,111 squared is 12345678987654321. The sum of digits is indeed 81. But why do some sources say 80? Let us check if there is a carry-over or if n=9 gives 80. Wait! For n=9, 111,111,111 × 111,111,111 = 123456787654321? No, let's check: 11^2 = 121 (sum 4 = 2^2). 111^2 = 12321 (sum 9 = 3^2). 1111^2 = 1234321 (sum 16 = 4^2). 11111^2 = 123454321 (sum 25 = 5^2). 111111^2 = 12345654321 (sum 36 = 6^2). 1111111^2 = 1234567654321 (sum 49 = 7^2). 11111111^2 = 123456787654321 (sum 64 = 8^2). 111111111^2 = 12345678987654321 (sum 81 = 81). Wait, why did 11111111^2 have 8 instead of 9? Because when n exceeds 9, carries occur. For n = 9, the middle digit is 9, so no carry exceeds 9. Thus the sum is 81. But wait, is option (c) 81 correct? Yes, 81 is option (c). Let us select index 2 (option C).