Classification – Odd one out (two-letter jumps): Identify the pair that is unlike the others: DH, FJ, HK, PR.

Introduction / Context:
Two-letter pair problems can be solved by checking consistent properties such as fixed step size or letter parity (odd/even alphabet positions). When step sizes vary among several options, a parity-based discriminator often cleanly isolates one outlier.

Given Data / Assumptions:

• Pairs: DH, FJ, HK, PR.
• Map letters to positions: A=1, B=2, …, Z=26.

Concept / Approach:
First, note that the forward step sizes differ (D→H is +4, F→J is +4, H→K is +3, P→R is +2), so step size alone does not produce a unique outlier. Instead, inspect parity of the second letter. If three pairs end at an even-indexed letter and one pair ends at an odd-indexed letter, the latter is the outlier.

Step-by-Step Solution:
DH: D(4)→H(8). Second letter H is position 8 (even).FJ: F(6)→J(10). Second letter J is position 10 (even).PR: P(16)→R(18). Second letter R is position 18 (even).HK: H(8)→K(11). Second letter K is position 11 (odd).

Verification / Alternative check:
Check first-letter parity: all first letters are even-indexed (D=4, F=6, H=8, P=16), so parity of the first does not discriminate. The second-letter parity provides the clean 3-to-1 split.

Why Other Options Are Wrong:

• DH, FJ, PR all terminate at even positions; only HK terminates at an odd position.

Common Pitfalls:
Assuming the rule must be a fixed jump size. Here, different jumps exist, so parity is the more reliable classifier.

Final Answer:
HK