2023 CSAT — Q75

Index the $20$ characters and slice into four blocks of length $5$ (positions $1$–$5$, $6$–$10$, $11$–$15$, $16$–$20$):

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

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

• Block 1 $= ABC\mathbf{DD}$ (double $D$) $\Rightarrow$ blanks @4,5 $= D, D$
• Block 2 $= AB\mathbf{CC}D$ (double $C$) $\Rightarrow$ blank @9 $= C$
• Block 3 $= A\mathbf{BB}CD$ (double $B$) — already complete, matches ✓
• Block 4 $= \mathbf{AA}BCD$ (double $A$) $\Rightarrow$ blank @16 $= A$

Reading the blanks in order (positions $4, 5, 9, 16$): $D,\ D,\ C,\ A$.

Answer: (d) DDCA.

$20 = 4 \times 5$, so the period is $5$, not $4$ — each block is $ABCD$ with exactly one doubled letter, and the doubled letter descends $D, C, B, A$ across the four blocks (the mirror of the 2024 Q24 ascending variant). Verify against the fully-printed block 3 ($ABBCD$, double $B$) to lock the direction, then the blanks fall out with zero ambiguity. The move that scores: count $20$, slice in $5$s, and read the doubled letter off the complete block.