IAS 2023 — CSAT Maths

Correct Option (C)

The problem involves multiplying a 9-digit number, where each digit is 1, by itself. This can be represented as $111,111,111 \times 111,111,111$111,111,111×111,111,111.

Observe the pattern for multiplying numbers consisting solely of the digit 1:

• $1 \times 1 = 1$1×1=1. The sum of digits is 1.
• $11 \times 11 = 121$11×11=121. The sum of digits is $1 + 2 + 1 = 4$1+2+1=4.
• $111 \times 111 = 12321$111×111=12321. The sum of digits is $1 + 2 + 3 + 2 + 1 = 9$1+2+3+2+1=9.

This pattern indicates that when a number consisting of 'n' ones is multiplied by itself, the resulting number is a palindrome that ascends from 1 to 'n' and then descends back to 1. For a 9-digit number (n=9) consisting of ones, the product will be:

$111111111 \times 111111111 = 12345678987654321$111111111×111111111=12345678987654321

To find the sum of the digits of this resulting number, we add each digit:

$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1$1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1

This sum can be calculated as the sum of integers from 1 to 9, plus the sum of integers from 1 to 8:

• Sum (1 to 9) = $\frac{9 \times (9+1)}{2} = \frac{9 \times 10}{2} = 45$29×(9+1)​=29×10​=45
• Sum (1 to 8) = $\frac{8 \times (8+1)}{2} = \frac{8 \times 9}{2} = 36$28×(8+1)​=28×9​=36

Total sum of digits = $45 + 36 = 81$45+36=81.

Incorrect Options:

Options (A) 64, (B) 80, and (D) 100 are incorrect because the calculated sum of the digits of the resulting number, based on the established mathematical pattern, is 81. These options do not correspond to the correct sum.