2024 CSAT — Q24

Index the 20 characters and locate the blanks before guessing — the pattern only reveals itself once the string is sliced into four equal blocks of length 5.

Positions $1$–$20$ carry blanks at $2,6,14,19$. Slice into blocks of $5$ (positions $1$–$5$, $6$–$10$, $11$–$15$, $16$–$20$):

• Block 1 ($1$–$5$): $A\ \_\ B\ C\ D$
• Block 2 ($6$–$10$): $\_\ B\ B\ C\ D$
• Block 3 ($11$–$15$): $A\ B\ C\ \_\ D$
• Block 4 ($16$–$20$): $A\ B\ C\ \_\ D$

Each block is the four letters $A,B,C,D$ with one letter doubled, and the doubled letter marches $A\to B\to C\to D$ across the blocks:

• Block 1 = $\mathbf{AA}BCD$ (double $A$) $\Rightarrow$ blank $@2 = A$
• Block 2 = $A\mathbf{BB}CD$ (double $B$) $\Rightarrow$ blank $@6 = A$
• Block 3 = $AB\mathbf{CC}D$ (double $C$) $\Rightarrow$ blank $@14 = C$
• Block 4 = $ABC\mathbf{DD}$ (double $D$) $\Rightarrow$ blank $@19 = D$

Every printed letter checks out (e.g. block 2 has $B,B,C,D$ at $7$–$10$ as given; block 3 has $A,B,C$ at $11$–$13$ as given). The four blanks, read in order, are $A,\ A,\ C,\ D$.

Answer: (c) A, A, C, D.

The whole item turns on finding the period. The early "$BCD$" and "$BBCD$" fragments tempt a length-$4$ read, but $20 = 4\times 5$, and only the length-$5$ slice (four blocks) exposes the rule: each block is $ABCD$ with the $k$-th block doubling the $k$-th letter ($AABCD,\ ABBCD,\ ABCCD,\ ABCDD$). Once you see the doubled letter “travelling” left-to-right, the blanks fall out with zero ambiguity. Under the clock, the move is: count the characters (20), test the divisor that is not the obvious 4, and read structure off the printed letters rather than the blanks.