CSAT Solved Papers/ 2025/Q48

2025 CSAT — Q48

Quant Number theory 2.5 marks Hard

What is the $489^{\text{th}}$ digit in the number $123456789101112\ldots$ ?

A 0
B 3
C 6
D 9 Answer

Worked rationale

The string concatenates $1, 2, 3, \dots$ . Count digits block by block (by number length):

$1$ -digit numbers $1$ – $9$ : $9$ numbers $\times 1 = 9$ digits (positions $1$ – $9$ ).
$2$ -digit numbers $10$ – $99$ : $90 \times 2 = 180$ digits (positions $10$ – $189$ ).

After $99$ we have used $189$ digits. The target is position $489$ , so we are $489 - 189 = 300$ digits into the $3$ -digit block (starting at $100$ ).

300 \div 3 = 100 \text{ exactly} \Rightarrow \text{the } 100^{\text{th}} \text{ three-digit number ends at position } 489.

The $100^{\text{th}}$ three-digit number is $100 + 99 = 199$ , and " $199$ " occupies positions $487, 488, 489$ . The $489^{\text{th}}$ digit is its last digit, $9$ .

Answer: (d) 9.

Why the other options miss

A

counted one too many: lands on the digit right after $199$ (the " $0$ " of $200$ at position $490$ ), overshooting by one position.
B

an arithmetic slip: mis-subtracts the cumulative count (e.g. uses $190$ instead of $189$ before the $3$ -digit block), shifting into the wrong number’s digit.
C

counted one too many: places $300/3 = 100$ as the $100^{\text{th}}$ number being $100$ (an off-by-one in indexing), reading a wrong digit.

Specialist insight

These “ $n$ -th digit of $123456789101112\dots$ ” items are pure block accounting: tally $9\cdot1, 90\cdot2, 900\cdot3, \dots$ digits and subtract until the remaining count lands inside a block. The two fence-post hazards are (i) the running total before the $3$ -digit block is $9 + 180 = 189$ , not $190$ , and (ii) the $k$ -th $3$ -digit number is $99 + k$ , not $100 + k$ . Here $300$ divides evenly by $3$ , so the position is the last digit of the $100^{\text{th}}$ three-digit number $199$ — a clean boundary that the off-by-one distractors (a)/(c) are built to catch. Carry the cumulative counts explicitly and index from the right block.

The trap, in one line

After $189$ digits ( $1$ – $99$ ), $300$ more $= 100$ three-digit numbers exactly, ending on the last digit of $199$ $\to$ $9$ ; one step off gives the " $0$ " of $200$ .

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