CSAT Solved Papers/ 2022/Q29

2022 CSAT — Q29

Quant Number theory 2.5 marks Medium

In the series $AABABCABCDABCDE\ldots$ , which letter appears at the $100$ th place?

A G
B H
C I Answer
D J

Worked rationale

The series is built from blocks: block $k$ is the first $k$ letters of the alphabet.

\underbrace{A}_{1} \mid \underbrace{AB}_{2} \mid \underbrace{ABC}_{3} \mid \underbrace{ABCD}_{4} \mid \underbrace{ABCDE}_{5}\ \cdots

Block $k$ has length $k$ , so the cumulative length after $n$ blocks is $\dfrac{n(n+1)}{2}$ .

Find the block holding position $100$ : after $13$ blocks, $\tfrac{13\cdot 14}{2} = 91$ ; after $14$ blocks, $\tfrac{14\cdot 15}{2} = 105$ . So position $100$ lies in block $14$ (positions $92$ – $105$ ).

Within block $14$ (which is $ABCDEFGHIJKLMN$ ), position $100$ is the $100 - 91 = 9$ th letter. The $9$ th letter of the alphabet is I.

Answer: (c) I.

Why the other options miss

A

counted one too few: places the block start at position $92$ but counts the offset as $7$ (e.g. uses cumulative $93$ ).
B

counted one too few: takes the offset as $8$ ( $100 - 92$ without the $+1$ ), landing one letter early.
D

counted one too many: takes the offset as $10$ , over-shooting by one.

Specialist insight

Two accountings decide it: which block ( $\tfrac{n(n+1)}{2}$ crosses $100$ at $n=14$ ) and the offset inside it ( $100 - 91 = 9$ , since block $14$ starts at position $92$ ). The fencepost is in the offset — the block begins at $92$ , so the $9$ th letter is position $100$ . Triangular-number block-length is the workhorse for every “ $k$ th term of a growing-block series.”

The trap, in one line

Cumulative $91$ after $13$ blocks; position $100$ is the $9$ th letter of block $14$ $= I \Rightarrow$ (c).

← All 2022 CSAT questions