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?

1. A. 64
2. B. 80
3. C. 81
4. D. 100

Answer: C

Explanation

Let the 9-digit number be R9 (a repunit of 9 ones), i.e., 111,111,111. We need to find the sum of the digits of R9 × R9.
Consider smaller repunits squared:
1^2 = 1 (Sum of digits = 1 = 1^2)
11^2 = 121 (Sum of digits = 1+2+1 = 4 = 2^2)
111^2 = 12321 (Sum of digits = 1+2+3+2+1 = 9 = 3^2)
1111^2 = 1234321 (Sum of digits = 1+2+3+4+3+2+1 = 16 = 4^2)
This pattern shows that for a number consisting of ‘n’ ones, its square results in a number whose digits sum to n^2. For a 9-digit number of all ones (n=9), the sum of the digits of its square will be 9^2 = 81.
Alternatively, a number consisting of ‘n’ ones is divisible by 9 if n is a multiple of 9. Here, n=9, so 111,111,111 is divisible by 9. Its square will also be divisible by 9. The sum of digits of any number divisible by 9 must also be divisible by 9. Among the options, only 81 is divisible by 9.