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?

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Accepted Answer

The correct answer is option C. 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.

Q18. 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?

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