CSAT Solved Papers/ 2023/Q28

2023 CSAT — Q28

Quant Number theory 2.5 marks Medium

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?

A 64
B 80
C 81 Answer
D 100

Worked rationale

The number is the repunit $R_9 = 111\,111\,111$ (nine $1$ s). Repunit squares follow a known palindromic pattern:

11^2 = 121,\quad 111^2 = 12321,\quad 1111^2 = 1234321,\ \dots

For $R_n$ with $n \le 9$ , $R_n^2$ reads $1\,2\,3\,\cdots(n-1)\,n\,(n-1)\cdots 2\,1$ . For $n = 9$ :

R_9^2 = 12\,345\,678\,987\,654\,321.

Sum its digits — it is $1{+}2{+}\cdots{+}8{+}9{+}8{+}\cdots{+}2{+}1$ :

= 9 + 2(1 + 2 + \cdots + 8) = 9 + 2\times 36 = 9 + 72 = 81.

Answer: (c) 81.

Why the other options miss

A

an arithmetic slip: sums only one side of the palindrome or stops the ascending run early ( $1+\cdots+8 = 36$ , doubled $= 72$ , forgetting the central $9$ ), or another partial total.
B

off by one: counts the central digit once but drops a $1$ at an end of the palindrome.
D

solved the wrong question: assumes the digit sum is $n^2 = 81$ confused with ” $10^2$ ,” or treats the square as having a different (non-palindromic) digit pattern.

Specialist insight

The pattern $R_n^2 = 123\cdots n \cdots 321$ holds only up to $n = 9$ (at $n = 10$ carries break the clean palindrome). Since $n = 9$ is exactly the boundary, the square is the full $12345678987654321$ and its digit sum is $9^2 = 81$ — equivalently $9 + 2\cdot\frac{8\cdot 9}{2}$ . Recognising the repunit-square palindrome turns a $9$ -digit multiplication into a one-line digit count; the slick check is “digit sum $= n^2$ for $R_n^2$ , $n\le 9$ .”

The trap, in one line

$R_9^2 = 12345678987654321$ (palindrome up to $9$ ); digit sum $= 9 + 2(1{+}\cdots{+}8) = 81 \Rightarrow$ (c).

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